@@PreMath: At 05:53, you wrote a^2 + b^2 = ±3*sqrt(13) . The negative value should be discarded because it's the sum of squares, a^2 + b^2 , which is non-negative for real a and b .
Correction: a^2 - b^2 = 9. Not like the shown. Sorry for the error. This engineer Greg calingasan 82.5 years old. Nagkakamalimali na kasi ako sa pagpindot.The problem given is really very easy to solve.
a^2 - b^2 = 9 ab = 3 b = 3/a a^2 - (3/a)^2 = 9 a^2 - (9/a^2) = 9 Let a^2 = x x - (9/x) = 9 Multiplying by x x^2 - 9 = 9x x^2 - 9x - 9 =0 Solve for x. Then a = square root of x Then b = 3/a
I used the same approach. After I simplified and rationalized, it stayed ugly. I got 1/2*(sqrt((9+3 sqrt(13))/2))*(sqrt(13)-1). Yep, that is the same answer though not as pretty.
This is the same approach I took, though in a modified form. I solved the quadratic x^2 -9x - 9 = 0 and rejected the negative root since x (=a^2) can't be negative. this give x = a^2 = [9+3sqrt(13)]/2. I did not use b=3/a to get the value of b but, instead, considered that b^2 = a^2 - 9. Thus a^2 + b^2 = 2a^2 - 9 = 9+3qrt13 - 9 = 3sqrt(13). From this, (a+b)^2 = a^2 + b^2 + 2ab = 6+3sqrt(13). We can not take the square root on both sides to get the value(s) for a+b).
@@nicholasjackson4941 your expansion of (a+b)^2 is incorrect. It does not equal a^2 -b^2 +2ab; rather, it is equal to a^2 +b^2 + 2ab. You must first find a value for a^2 + b^2 (which turns out to be 3sqrt(13)).
I think this is a better approach: notice that 9=3^2 = (ab)^2. The first equation can be rewritten as: a^2-b^2 = (ab)^2, we get a quadratic equation of (a/b). Solve this equation and together with ab=3, it's easy to find the value of a and b and consequently a+b.
From the second equation, b = 3/a. Substitute the value of b into the first equation: a² + (3/a)² = 9. Multiply both sides by a² to get a⁴ + 9 = 9a². Rearrange to get a⁴ - 9a² +9 = 0. Let x = a² to make this a quadratic: x² -9x +9 = 0. Solve with quadratic formula x = (9 + √(81 + (4)(1)(9))/2 or x = (9 - √(81 + (4)(1)(9))/2 simplifying to x = (9 + 3√(13))/2 or x = (9 - 3√(13))/2. Replace x by a² and note that the second value of x is negative. Ruling out imaginary numbers for a, x must be positive, a² = (9 + 3√(13))/2. From the first equation, b² = ((9 + 3√(13))/2) - 9 which simplifies to b² = (3√(13)-9)/2. we note that, for ab = 9, a and b must both be positive or both negative. Taking the positive roots, a + b = √((9 + 3√(13))/2) + √((3√(13)-9)/2). Using a scientific calculator, a = 3.147749500 and b = 0.953061862. a + b = 4.100811362. PreMath's solution of a + b = 6 + 3√(13) = 4.100811362 is identical when calculated to the same level of precision. I suspect that my solution can be simplified to be identical to PreMath's solution. Taking negative values for my a and b produces a solution with the same magnitude but a negative sign, PreMath's other solution.
Nice, I got exactly the same answer, the one with the sum of two roots. Actually, there is a way to simplify the sum to the answer from video. A well-known formula sqrt(a±bsqrt(c)) = sqrt((a+sqrt(a²-b²c))/2) ± sqrt((a-sqrt(a²-b²c))/2) is valid for positive real a, b, c such that a² - b²c >= 0. But if we take a, b, c with a² - b²c < 0 condition, we'll get complex values under the outer roots. However, one can proof that for a² - b²c < 0 the formula looks like sqrt(±a+bsqrt(c)) = sqrt((bsqrt(c)+sqrt(b²c-a²))/2) ± sqrt((bsqrt(c)-sqrt(b²c-a²))/2) where a, b, c > 0. In our example a = 9, b = 3, c = 13: sqrt(±9+3sqrt(13)) = sqrt((3sqrt(13)+ 6)/2) ± sqrt((3sqrt(13)-6)/2) Therefore, their sum is 2sqrt((3sqrt(13)+ 6)/2) = sqrt(6sqrt(13)+ 12). And finally, dividing by sqrt(2) will give you the answer from video
An easier way is to use a right angles triangle with a as hypotenuse and the other 2 sides as b and 3. This uses the given info which is a^2 - b^2 = 9. Use the second info ab = 3 to get b = 3/a. Now use the right angle triangle to get a^2 - (3/a)^2 = 9. In the calculation use a substitution x = a^2. Now solve a quadratic eqn to get x and then solve for a and b. Now do a + b. Very very simple. Unlike the method used in this video. Thanks. Have a great day.
Indeed. Or simply substitute a = 3/b in (a+b)(a-b)=9. Substitute a squared for d, then do the same with b = 3/a and substitute b squared for e. Solve quadratic equations for d and e using formula. Now use identity (a + b) all squared = d + e + 2ab, substituting with the values calculated above. I feel this was quicker and simpler.
@@VolksdeutscheSS hi Dear. I'm a teacher of Maths for over 49 years. I know confusing students with alternate tedious methods is unfair. So you need to be thankful for people like us that make the work load of students appreciative of us humble professions also. I normally do engage more able students to offer alternate math solutions.
We can still simplify the problem in the following manner. Let x = a² and y = b². So that x - y = a² - b² = 9 ... (1). Again xy = a².b²=(ab) ² = 3² = 9 ...(2) Now using the algebraic equation (x + y )² = (x - y )² + 4xy , we get (x + y )² = 9² + 4. 9 = 117. So that (x + y ) = √117. Here negative sign is not considered as x and y are square terms of a and b . Replacing the values of a and b for x ad y, we get a² + b² = √117. Now (a + b) ² = (a² + b²) + 2ab = √117 + 2. 3 = √117+6. Therefore a + b = +-√ (√117+6), thus our result.
A shorter solution: Substitute x=a+b, y=a-b. We obtain from here that xy=9, x^2-y^2=12. We can remove y from this system to obtain a biquadratic equation for x
@@EfSaNe2531 In a way it is more straightforward, since we're interested in x=a+b, not a or b in itself. The other thing to note is the use of binomic formulas, the 3rd occurs obviously in the 1st eqn of the problem, the 1st and 2nd more indirectly as (a+b)^2-(a-b)^2=a^2+2ab+b^2-a^2+2ab-b^2=4ab, which connects with the 2nd eqn of the problem. I.e. after recognizing the 3rd binomial equation and noticing that we're really interested in a+b, not a or b in itself, it makes sense to change variables to x=a+b, y=a-b and see how the second equation of the problem can be expressed in x and y, then eliminate y. That solution path always has the solution in its view. The video solution has a lot of steps that only make sense later. E.g. why calculate a^4+b^4? Only when you already know where you're going next that makes sense. And even that is then first leading to a^2+b^2, another intermediate step where you have to know how that is going to help you.
@@jootpepet Yes, it's a kind of doing math like a stage magician: lots of handwaving (term transformations), then the solution comes out of nowhere. Doing lots of tricks without really explaining how it is done. Of course each transformation in itself works out, but it isn't motivated, why it is done. Of course sometimes one way to arrive at a solution is to play around with the equations, and in the end pick the way that works, but it's kind of stumbling around in a labyrinth, then presenting the way you found without ever mentioning all the dead ends.
@@Pengochan sir this is about practice. No-one can teach u shorter solutions in short time but u urself had to do practice so that those short tricks come in mind at right time. If u want explanation then it is " as question is asking to find a+b instead of a or b . Just rearrange both equation in form of a+b and a-b." This method will come in ur mind if u have done practice otherwise it won't come until unless u r too smart who don't need practice.
@@ichvvhovxugcvuhi3522 If a or b can be complex numbers, then at 8:00 still cannot be supposed that (a + b)^2 >= 0. If a and b are supposed to be real numbers, in the video he can reject the wrong negative solution at 6:00 instead of 8:00 as @gold user stated.
@@chessandmathguy but the question didn't specify. Better to be inclusive of all possible answers real or complex. When the question specify other conditions then reject the respective value(s).
You can go directly from a^4 - 2a^2b^2 + b^4 = 81 to a^4 + 2a^2b^2 + b^4 = 81 + 4a^2b^2. The LHS is (a^2 + b^2)2 and the RHS is 117. So instead of adding 18 twice you just add 36 once. And you don't need to consider a^2 + b^2 = - 3 sqrt(13). You keep that option around far longer than necessary.
consider the system (1) a ^ 2 + b ^ 2 = k (where ^ 2 stands for squared) a ^ 2 - b ^ 2 = 9 from which adding and subtracting the two equations member by member a ^ 2 = (k + 9) / 2 and b ^ 2 = (k - 9) / 2 (2) a ^ 2 * b ^ 2 = (ab) ^ 2 = 3 ^ 2 = 9, applying the (2) k ^ 2 - 81 = 36 k ^ 2 = 117 = 13 * 9 and k = ± 3 √13 from (a ^ 2 + b ^ 2)> 0, k = + 3√13 follows a ^ 2 + b ^ 2 = (a + b) ^ 2 -2ab = + 3√13 and (a + b) ^ 2 = 6 + 3√13 (the solution 6 - 3√13 must be discarded because it is negative in contrast to the first member of equality which is always positive) a + b = ± √ (6 + 3√13)
Actually, this is a very simple and trivial problem that can be solved directly by expressing unknown a (and b) from the second equation ab=3, and by substituting it to the first equation. In this obvious method, we get a quadratic equation for a^2 (and for b^2). Having received a^2 and b^2, it is very easy to find (a+b)^2 and solve the problem. Such a straightforward solution does not require any tricky actions (as it is done in this video). In my opinion, this problem does not correspond to the level of Math Olympiad, since it is very simple, and it can be done to the students in any regular (not math) class as a homework. BTW, since we are dealing with real numbers only (not complex), it’s absurd to use a minus sign in the expression (a^2+b^2) because it can't be negative.
@@trexdantea9458 I think the host of the channel is trying to teach people substitution algebra. All his videos use this method. That's why it's used here even though there's a simpler way to do it.
In an examination this question carries 1 or 2 marks. This method would be too long. In my own view just make a or b the subject of the formula in the second eqn and substitute in the first eqn. Get quadratic equation. Solve the period
Since a^2 - b^2 = 9 and ab = 3 then we can write a^2 - b^2 = ( a-b) (a+b) = 9 that means a + b = 9 / a - b if we substitute a = b/3 in that equation to becomes 3/b + b = 9 / (3/b - b) then by solving this equation we will get the following form : b^4 + 9b^2 - 9 = 0 now let's call b^2 = x then we get x^2 + 9x - 9 = 0 and by solving this equation we will get x = 0.908 and we know that x = b^2 that means b = 0.953 then a = 3, 1477 and a + b = 4,1 and that is the same answer. Thank you very much sir.
That is basically what I did to solve the equation. I used the letter "u" instead of "x" and solved for a instead of b. I ended up with a^4 -9a^2 -9 and plugged in u for a^2, u^2 - 9u - 9 = 0. Solving gives the approximate values of u = 9.908327 and u = -0.953062. Substituting back for a^2 means a^2 = 9.908327 and a^2 = -0.953062. a = +/- 3.147750 and a = +/-0.953062 * i (which is rejected), so a = +/- 3.147750. If b = 3/a, then b = +/- 0.953062. (a + b) is approximately +/- 4.100812.
I suppose it would be a lot easier if you use the concept a square-b.square= (a+b)(a-b)=9 and it gives a+b =9/a-b, given the ab =3, a+b =9/a-a/3, a+b=27/2a, giving a.square=81/2, giving a=9/root2 ,b now=3/a=3xroot2/9=root2/3; therefore.a+b= 9/root2+root2/3, a=b =29xroot2/3
Set t := a + b. Because (a-b)^2 = t^2 - 4ab = t^2 - 12, we have 81 = (a^2-b^2) = (a-b)^2(a+b)^2 = t^2(t^2-12), or equivalently t^4 - 12t^2 -81 = 0. Calculating the positive root of x^2 - 12x - 81 = 0 and taking ±sqrt of it, we obtain the values of t.
The most important thing is...We know that {ab = 3} and{ a^2-b^2 = 9} means (a+b)(a-b) = 9. and What we are finding is (a+b) NOT (a, b). better to ignore each of a, b. At this point, the best approach is... Just Put X = (a+b),Y = (a-b) and... use Basics of math, below --> DONE. [1]. (a+b)^2 = a^2+b^2 + 2ab [2]. (a-b)^2 = a^2+b^2 - 2ab Obviaously, [2] = [1] + 4ab ,and we know (ab=3). [3]. Y^2 = X^2 + 12 [4]. XY=9 with [3] and [4], we get quadro of func(X), and find X^2, and find X. (still skipping each of a , b).
Hi there and thanks for this nice system. Another shorter solution is to take X=(a+b)^2 and Y=(a-b)^2 . We can easily get that X-Y=12 and XY=81. It’s then easy to find X and then a+b
Let a + b = c. By an express from this equality `a`(b) and squaring both sides we are getting a^2 - b^2 = c^2 - 2ac and b^2 - a^2 = c^2 - 2bc. Hereof c^2 - 9 = 2bc and c^2 + 9 = 2ac, from product of the last two equations are getting c^4 -12 c^2 - 81 = 0. By solved this biquadratic equation we are getting the same solution.
I have seen some of videos. How do you come up with a strategy for solving these equations? The connection from the start to the end is not obvious. What kind of intuition to you use?
Great video as always, found it even trickier than most of your other videos 😁 I am wondering: Could you solve a task like this one quickly if you were asked? I think it is anything but straight forward.
Thank you for your feedback! Cheers! These kind of problems require slick moves, consistent practice, and continued exposure. No wonder I reveal all kind of question for my audience that require various techniques... You are awesome, Tobias. Keep it up 😀
This one stretched me further than most of your other videos, so I will have to go through it again more slowly. I got part way without watching the video first, but I think remembering those identities is important for me.
Fulgurant résolution : Aristarchus identity : (a^2+b^2)^2 = (2ab)^2 + (a^2-b^2)^2 Thus : (a+b)^2 = 2ab +(a^2+b^2)= 2ab + sqrt[(2ab)^2 + (a^2-b^2)^2]. Which gives immediately the two general solutions for a+b. For particular solutions use : ab=3 and (a^2-b^2)=9 to get : (a+b)^2 = 6+3sqrt(13) QED
Simple and standard way to solve this is: subtracting the second equation multiplied by 3 from the first, we get: a^2 - 3ab - b^2 = 0, This is homogeneous equation - sum of all powers is 2. And then b = k * a - is standard substitution for homogeneous equation. a^2 - 3 a^2 k - a^2 * k^2 = 0; a 0; k^2 + 3k - 1 = 0; k = (-3 +- sqrt(13) / 2; and ab = 3 , hence a^2 * k = 3, hence k > 0, and the only k = (-3 + sqrt(13)) / 2. and then from a^2 * k = 3, a = +- sqrt( (3/2) * (sqrt(13) + 3) ), and from b = k*a b = +- sqrt( (3/2) * (sqrt(13) - 3) ).
consider the system (1) a squared + b squared = k a squared - b squared = 9 a ^ 2 or b ^ 2 = a squared or b squared from which adding and subtracting the two equations member by member a ^ 2 = (k + 9) / 2 ... b ^ 2 = (k - 9) / 2 ... (2) a ^ 2 * b ^ 2 = (ab) 2 = 32 = 9, applying (2) k ^ 2 - 81 = 36 ... k ^ 2 = 117 = 13 * 9 and k = ± 3 √13 therefore a ^ 2 + b ^ 2 = + 3√13 (sum of two squares necessarily positive) a ^ 2 + b ^ 2 = (a + b) ^ 2 -2ab = 3√13 ... (a + b) ^ 2 = 2ab + 3√13. (a + b) ^ 2 = 6 + 3√13 a + b = ± √ (6 + 3√13)
I am trying to figure out how to do a simple math problem, my daughters teacher couldn't help. If I buy a stock for $30 dollars and it goes down to $28.75 and I end up losing $250. I am trying to find out how many shares I had in order to lose $250? I am trying to figure out what steps I have to do In order to know how many shares of a stock I need to buy In order to lose the same amount on different trades. Thanks for any responses.
There is a much more general, systematic, elegant, concise and faster way (which is quite surely the optimum), by use of the Aristarchus identity, whose modern version is the complex number quadrance (or modulus) fundamental theorem : Q(z)Q(Z) = Q(zZ) Where z=a+ib, Z=A+iB, and Q(z) = a^2+b^2 is the quadrance of the complex number z. Indeed in the special case z=Z it simplifies to : (a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2 From that starting point the general solution to the problem follows immediately : (a+b)^2 = 2ab +(a^2+b^2)= 2ab + sqrt[(2ab)^2 + (a^2-b^2)^2]. (the negative sqrt is excluded by positivity of the LHS) Which gives the two solutions for a+b. QED
I may have done an equivalent thing as you describe (but without really being sure if it's legitimate or not--though it does give the numerically correct answer). Basically, completing the square with complex numbers. Multiply eqn (2) by 2i and add to eqn(1). Then you get a^2 +2abi - b^2 = 9 +6i. The left hand side simplifies to (a+bi)^2. Then take the complex square root of the right hand side; a and b are the real and imaginary parts of the result.
@@bradwilliams7198 No! All what you've writen is perfectly correct, but unfortunately you are making a CIRCULAR reasoning, that you don't see yet, since you didn't finish your calculation. Indeed, you get to the edge point of computing the square root of 9+6i. But how do you actually compute it?... Obviously by looking for two real numbers a and b which satisfy the equation : (a+bi)^2 = 9+6i. And then what? Well you have little option else than developing this identity, which leads you to...your starting point? Isn't it so?... 😉 That's why you made a CIRCULAR reasoning, hidden in you unfinished resolution. Why is that hapening ? Because you only rewrite the problem in an other form, without actually decreasing it's "entropy". You didn't actually FACTORISE, even thow you have made a factorisation. But the one you're making, though correct, DOESN'T lift the problem to a lesser "entropic" level, which is only done by Aristarchus identity that I expose. Such identities are corner stones. You canot avoid them. They are the keys of algebraic properties, like the very EXISTENCE of "complex numbers". In such way that it's the lack of existence of a similar indentity in 3D, that prevent a "complex structure" to existe in 3D. And so on in 5D, 6D, 7D, 9D, etc. Only in 4D and 8D, such identities exist, similar to Aristarchus one in 2D. In other words, though they may seem to be several ways to solve the problem. There is in fact only one! All correct ways will necessarely use in a way or another the Aristarchus identity, which is THE door keeper of 2D world. Furthermore, I didn't actually make any use of complex number in my resolution. I simply apply directly Aristarchus indentity which gives almost immediately, with hardly any computing, the two opposit solutions : (a+b)^2 = 6+3√13. My mention of complex number was only to recall one of the fastest, concise and easy to remember computable way to derive Aristarchus identity. But such identity was discover thousands of years before the discovery of complex numbers by Tartaglia, Cardano, Ferrari, Bombelli,...in the XVI th century. The identity holds by itself already on pure Real numbers. Nevertheless it opened the door, long before their discovery, to the very EXISTENCE of this for long hidden multiplicative structure on 2D vectors, that we call "complex numbers". Thus to conclure this short glance in Mathematics History and hidden Arcanes, the very existence, or lack of existence, of such corner stones algebraic identities, tells us about the somehow MORPHOLOGY of spaces, in different dimensions. In other words, it is highly non trivial nor innocent, to add a dimension... It may change a hell of things! In 4D, though the QUATERNIONS exist, thanks to a similar Real algebraic 4D identity, COMMUTATIVITY nevertheless is LOST! And furthermore in 8D, where the existence of OCTONIONS arises thanks to, again, the "miraculous" existence of a similar Real algebraic identity, it's even worse than with the QUATERNIONS, since not only COMMUTATIVITY is lost, but also the important arithmetic ASSOCIATIVITY. Which starts to bring OCTONIONS at the slipery edge of USEFUL numbers world!
@@Igdrazil Well, the easiest way to take the square root of a complex number (e.g. a +bi) is to convert to polar coordinates r = (a^2 + b^2)^0.5 ; theta = arctan(b/a). Then the square root will be (in polar coordinates) r^0.5, theta/2 (or theta/2 + pi for the second solution). Then convert back to cartesian coordinates. So (9+6i)^0.5 = 117^0.25*(cos(0.5*arctan(6/9)) +i*sin(0.5*arctan(6/9)). Numerically, 3.14774... + i * 0.95306...
Write X = a+b, such that a-b=9/X, using (a+b)(a-b)=9. Obviously (a+b)^2 - (a-b)^2 = 4ab, and thus X^2 - 81/X^2 = 12, such that (X^2)^2 - 12*X^2 - 81 = 0. Solving this quadratic equation gives X^2 = 6 + sqrt(117), and thus a+b = X equal to sqrt(6 + sqrt(117)) and -sqrt(6 + sqrt(117)).
There is a more simple solution: From handling Pythagorean triples we know, that a, b, c in an rectangular triangle may be represented by the expressions m^2 - n^2, 2mn and m^2 + n^2 for a and b, c respectively. Let x, y and z be the 3 sides of an rectagular triangle. So if a^2 - b^2 = 9 = x and ab = 3, y has to be 2ab = 6. Thus, z will be 3*sqrt(13). So (a+b)^2 = a^2 + b^2 + 2ab is equal to 3*sqrt(13) + 6 and ab = +/-sqrt(6 + 3*sqrt(13)).
At the end of Step 1 you had a^4 + b^4 =99. It seems like you should have been able to take the 4th root of both sides and ended up with a + b = 4√99. But 4√99 does not equal +/-2√(6 + 3√13).
i think i've found a better way to solve, we can use quadratic equations: M = A-B S = A+B delta = b^2 - 4ac => M = sqrt(b^2 - 4ac) / |a| => in this case we divided the whole equation by a, so a = 1 and equation is: P = A*B = 3 => x^2 - Sx + 3 = 0 and delta = S^2 - 12 so since a = 1 => M = sqrt(S^2 - 12) (a+b) (a-b) = a^2 - b^2 = 9 S * M = 9 S * sqrt(S^2 - 12) = 9 sqrt(S^2 - 12) = 9 / S S^2 - 12 = 9^2 / S^2 => S^2 = t t - 12 = 81 / t t^2 - 12t - 81 = 0 t = 6 + 3 * sqrt(13) => S^2 = 6 + 3*sqrt(13) => S = (A+B) = sqrt(6 + 3*sqrt(13)) ~~t = 6 - 3 * sqrt(13) => since 3 * sqrt(13) > 6, so this is not acceptable~~
My idea was different from this. I had some idea 1)a,b>0 2)a,b0 a/b=(3+sqrt13)/2 b/a=2/(3+sqrt13) sqrt a/sqrt b +sqrt b/sqrt a= (a+b)/sqrt 3 But my Renault for a,b>0 Is (sqrt 3)*{(5+sqrt13)/[(sqrt2)(sqrt(3+sqrt13)] Its the samé on calculator, but not so nice .....
a² - b² = 9 (1) ab = 3 (2) Lets divide (1) on (2) a/b - b/a = 3 (3) x - 1/x = 3 (4) x² - 3x -1 = 0 standard square equation with solutions m so ab = 3 and a/b = m a+b = √(3m) + √(3/m) where m = (3 + √13)/2 Notes : can't use second root (3 - √13)/2 since it's negative. Alternatevely, a + b = √(a² + 2ab + b²), so a+b = √(3m + 6 + 3/m)
em....I think the question should mentioned that a and b is real number at the very beginning. If it state a and b is natural number instead, then obviously it is no solution ab = 3 imply both a and b is odd number (a+b)(a-b) = 9 which doesn't means sense as both a+b and a-b can only be even number, product of 2 even numbers can only be even number, so a+b is no solution
Step 2. -(a^2+b^2+2ab) = -a^2-b^2-2ab. It does not = +a^2+b^2+2ab (I put + sign in front to emphasize that it is positive) In other words, multiplying by -1 as you did in step 2 causes the a coefficient to be negative, where in step 1 it is positive. Hope I could help! 😁😁
Thanks for solving this. First, l want to say this solution is perfect but l used another algebraic formula for this question. I used a⁴+b⁴=((a+b)²-2(ab))²-2a²b². And when l watched this solution l saw it is better.👍
if I wrote down: +- square root of (4.5 + square root of 29.25) + 3 : +-square root of (4.5 + square root of 29.25) does that count as the correct answer? The numerical value of this expression is the same as the value of the expression that the author of the video received, it's just that my record is longer (if I didn't have to write the square root in words all the time, it would be shorter). Sorry for my english, it's not my native language
I think this is shorter from equation (2) ab = 3 then we get a2 b2 = 9 then substitute in equation (1) a2 - b2 = a2 b2 a2 - b2 - a2 b2 =0 a2(1-b2) - b2 = 0 (3) form equation (2) we get a=3/b or a2=9/b2 then substitute in equation (3) (9/b2)(1-b2) -b2 =0 let b2 = y then we get 9(1-y)/y - y = 0 , multiply by y we get 9-9y-y2=0 (4) y=(9 [+,-] Sqr(81 - (4 * 9 * -1) ))/(2*9) y=(9 [+,-]Sqr(117))/ 18 we know that y must be positive and Sqr(117)>9 then y=(9+Sqr(117))/18 = 1.100925213 b = [+,-] Sqr(y) =[+,-] Sqr(1.100925213) = [+,-] 1.0492 a = 3/b =[+,-] 3 /1.0492 = [+,-] 2.8592 a+b = [+,-] (2.8592 + 1.0492) = [+,-] 3.9084
The path you expose here, toward the correct solution, is good. Nevertheless a bit lengthy with furthermore some extra missed shortcuts like your lack of simplification with a2+b2=3sqrt13. But remarquably, there is a much more general, systematic, concise and faster way (which is very probably the optimum), by use of the Aristarchus identity (which modern form is the complex number quadrance (or modulus) fundamental theorem) : Q(z)Q(Z) = Q(zZ) Where z=a+ib, Z=A+iB, and Q(z) = a^2+b^2 is the quadrance of the complex number z. Indeed in the special case z=Z it simplifies to : (a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2 From that starting point the general solution to the problem follows immediately : (a+b)^2 = (a^2+b^2) + 2ab = 2ab + sqrt[(2ab)^2 + (a^2-b^2)^2]. (the negative sqrt is excluded by positivity of the LHS) Which gives the two solutions for a+b QED
The best way for you please trying solving this task using complex analysis method, then you Will finding imaginary and real component and then those shall be completed to solve
This problem is a little harder than it looks. My solution is fairly similar to professor's. First take (a-b)*(a+b)=a^2-b^2=9 (equation 3). Square this to get (a-b)^2 * (a+b)^2 = 81 (equation 3a). Next note that (a+b)^2-(a-b)^2= (a+b)^2-(a-b)^2=4ab=12 (equation 4). Rearrange this to (a-b)^2=(a+b)^2-12 (equation 4a). We can now get rid of the (a-b)^2 by inserting equation 4a into 3a. The result is [(a+b)^2-12]*(a+b)^2)=81. (equation 5). Now define u=(a+b)^2. Equation 5 then becomes (u-12)*u=81 (equation 5b) Or u^2-12u-81=0 (equation 5c). The solution to this quadratic is u=[12+-sqrt(468)]/2. Or u = 6+-3sqrt(13). Since u is a square and thus must be positive we reject the negative root 6-3sqrt(13), thus u=6+3sqrt(13). Since u = (a+b)^2, then a+b=+-sqrt[3sqrt(13)].
There is an unstated assumption that the solution must be in real numbers. If you allow complex numbers, then you cannot say that (a+b)^2 = 6-3(sqrt(13)) is a false statement. This assumption should be included in the problem statement.
sqrt both sides a-b=3 a-3=b ab=3 (a)a-3=3 used a calculator and got sqrt6 for a (sqrt6)^2-(sqrt6 -3)^2 6-6+9 6-6=0 0+9=9 a=sqrt6 b=sqrt6-3 I think this is right, but I'm about to go into highschool next year so I wouldn't be to surprised if it was wrong, is it?
I still do not understand why the second solution was rejected... there are lots of cases when the square of a number is a negative number. No problem.
Thanks for the visit, Vectoria You're welcome! We don't have a book. We just do constant search and planning. If I find any good book, I'd love to share with you. Thanks for asking! Cheers! You are awesome. Keep it up 😀
It is assumed that a and b are real numbers. 9²=(a²-b²)²=(a+b)²(a-b)²=(a²+b²+2ab)(a²+b²-2ab)=(a²+b²)²-4(ab)²=(a²+b²)²-4x3² ∴ (a²+b²)²=81+36=117. ∴a²+b²=√117=3√13. (a+b)²=a²+b²+2ab=3√13+6. ∴a+b=±√(6+3√13).
Yes you have used in fact a simplified form of the Aristarchus identity which modern version is the complex number quadrance theorem : Q(z)Q(Z) = Q(zZ)
Okay, so I introduced new variables; p = a + b and m = a - b. p*m = (a + b)(a - b) = a^2 - b^2 = 9 and p^2 - m^2 = (a + b)^2 - (a - b)^2 = 4ab = 4 * 3 = 12. So I switched to a system with p and m, where I need to find p. p * m = 9 --> m = 9/p --> m^2 = 81/p^2. Substitute that in the second equation: p^2 - 81/p^2 = 12. Multiply by p^2 to get p^4 - 81 = 12 * p^2. Move everything to the same side and we get p^4 - 12 * p^2 - 81 = 0. This is a biquadratic equation, so we can solve for p^2 with the quadratic formula: p^2 = [-(-12) + - sqrt((-12)^2 - 4 * 1 * (-81))]/2 = [12 + - sqrt(468)]/2 = [12 + - 6*sqrt(13)]/2 = 6 + - 3*sqrt(13). 3*sqrt(13) > 6, so that cannot be an answer to p^2 (if we ignore complex answers). So p^2 = 6 + 3*sqrt(13). Which leads to p = a + b = + - sqrt(6 + 3*sqrt(13)).
Let X = (a+b)^2 = a^2 +2ab +b^2 Then summing this with the first equation leads to 2a^2 + 2ab = X + 9 hence 2a^2 = X + 3 [ab = 3] Substraction gives 2b^2 + 2ab = X - 9 hence 2b^2 = X - 15 Let's multiply : 2a^2 x 2b^2 = (X+3)(X-15) = 4 x (ab)^2 = 36 which leads to X^2 - 12X - 45 = 36 X^2 - 12X - 81 = 0 Delta is 12^2 + 4x81 = (2^2 x 2^2 x 3^2) + (2^2 x 3^2 x 3^2) = 6^2 x 13 Sqrt(Delta) = 6Sqrt(2^2+3^2) = 6Sqrt(13) X is a square so > 0, only positive root is (12 + 6Sqrt(13)) / 2 = 6 + 3Sqrt(13) finally a+b = Sqrt(6 + 3 Sqrt(13))
Great. Thank you.
You are very welcome.
Glad to hear that!
Thank you for your feedback! Cheers!
You are awesome, Luigi 😀
A
You can also watch my maths channel. @infostopiq
@@PreMath: At 05:53, you wrote a^2 + b^2 = ±3*sqrt(13) .
The negative value should be discarded because it's the sum of squares, a^2 + b^2 , which is non-negative for real a and b .
Correction: a^2 - b^2 = 9. Not like the shown. Sorry for the error. This engineer Greg calingasan 82.5 years old. Nagkakamalimali na kasi ako sa pagpindot.The problem given is really very easy to solve.
a^2 - b^2 = 9
ab = 3
b = 3/a
a^2 - (3/a)^2 = 9
a^2 - (9/a^2) = 9
Let a^2 = x
x - (9/x) = 9
Multiplying by x
x^2 - 9 = 9x
x^2 - 9x - 9 =0
Solve for x.
Then a = square root of x
Then b = 3/a
I used the same approach. After I simplified and rationalized, it stayed ugly. I got 1/2*(sqrt((9+3 sqrt(13))/2))*(sqrt(13)-1). Yep, that is the same answer though not as pretty.
This is the same approach I took, though in a modified form. I solved the quadratic x^2 -9x - 9 = 0 and rejected the negative root since x (=a^2) can't be negative. this give x = a^2 = [9+3sqrt(13)]/2. I did not use b=3/a to get the value of b but, instead, considered that b^2 = a^2 - 9. Thus a^2 + b^2 = 2a^2 - 9 = 9+3qrt13 - 9 = 3sqrt(13). From this, (a+b)^2 = a^2 + b^2 + 2ab = 6+3sqrt(13). We can not take the square root on both sides to get the value(s) for a+b).
(a+b)^2=a^2+2ab-b^2
=a^2-b^2+2ab
= 9+2(3)
=15
a+b=square root of 15
What is wrong in this solving please
@@nicholasjackson4941 your expansion of (a+b)^2 is incorrect. It does not equal a^2 -b^2 +2ab; rather, it is equal to a^2 +b^2 + 2ab. You must first find a value for a^2 + b^2 (which turns out to be 3sqrt(13)).
@@nicholasjackson4941 (a+b)^2 = a^2 + b^2 + 2ab
I think this is a better approach: notice that 9=3^2 = (ab)^2. The first equation can be rewritten as: a^2-b^2 = (ab)^2, we get a quadratic equation of (a/b). Solve this equation and together with ab=3, it's easy to find the value of a and b and consequently a+b.
Nice approach. But your right hand side should read 3*ab, to make your argument work.
@@olerask2457 Wow, you are so right. I totally missed this point and was actually wrong about the original problem. Thanks!
From the second equation, b = 3/a. Substitute the value of b into the first equation: a² + (3/a)² = 9. Multiply both sides by a² to get a⁴ + 9 = 9a². Rearrange to get a⁴ - 9a² +9 = 0. Let x = a² to make this a quadratic: x² -9x +9 = 0. Solve with quadratic formula x = (9 + √(81 + (4)(1)(9))/2 or x = (9 - √(81 + (4)(1)(9))/2 simplifying to x = (9 + 3√(13))/2 or x = (9 - 3√(13))/2. Replace x by a² and note that the second value of x is negative. Ruling out imaginary numbers for a, x must be positive, a² = (9 + 3√(13))/2. From the first equation, b² = ((9 + 3√(13))/2) - 9 which simplifies to b² = (3√(13)-9)/2. we note that, for ab = 9, a and b must both be positive or both negative. Taking the positive roots, a + b = √((9 + 3√(13))/2) + √((3√(13)-9)/2). Using a scientific calculator, a = 3.147749500 and b = 0.953061862. a + b = 4.100811362. PreMath's solution of a + b = 6 + 3√(13) = 4.100811362 is identical when calculated to the same level of precision. I suspect that my solution can be simplified to be identical to PreMath's solution.
Taking negative values for my a and b produces a solution with the same magnitude but a negative sign, PreMath's other solution.
I literally thought exactly like this.. thank you.
Nice, I got exactly the same answer, the one with the sum of two roots.
Actually, there is a way to simplify the sum to the answer from video. A well-known formula
sqrt(a±bsqrt(c)) = sqrt((a+sqrt(a²-b²c))/2) ± sqrt((a-sqrt(a²-b²c))/2)
is valid for positive real a, b, c such that a² - b²c >= 0. But if we take a, b, c with a² - b²c < 0 condition, we'll get complex values under the outer roots. However, one can proof that for a² - b²c < 0 the formula looks like
sqrt(±a+bsqrt(c)) = sqrt((bsqrt(c)+sqrt(b²c-a²))/2) ± sqrt((bsqrt(c)-sqrt(b²c-a²))/2)
where a, b, c > 0.
In our example a = 9, b = 3, c = 13:
sqrt(±9+3sqrt(13)) =
sqrt((3sqrt(13)+ 6)/2) ± sqrt((3sqrt(13)-6)/2)
Therefore, their sum is 2sqrt((3sqrt(13)+ 6)/2) = sqrt(6sqrt(13)+ 12). And finally, dividing by sqrt(2) will give you the answer from video
The last thing that I remember is that I was looking for a song....
Happens to me a lot. I wonder if it's contagious?
you hunt for a song but you maybe didn’t specify your search criteria lol.
An easier way is to use a right angles triangle with a as hypotenuse and the other 2 sides as b and 3. This uses the given info which is a^2 - b^2 = 9. Use
the second info ab = 3 to get b = 3/a. Now use the right angle triangle to get a^2 - (3/a)^2 = 9. In the calculation use a substitution x = a^2. Now solve a quadratic eqn to get x and then solve for a and b. Now do a + b. Very very simple. Unlike the method used in this video. Thanks. Have a great day.
Indeed. Or simply substitute a = 3/b in (a+b)(a-b)=9. Substitute a squared for d, then do the same with b = 3/a and substitute b squared for e. Solve quadratic equations for d and e using formula. Now use identity (a + b) all squared = d + e + 2ab, substituting with the values calculated above. I feel this was quicker and simpler.
@Siva: You don't have to be so "snippy". LOL. I like this channel. PreMath is just showing us another method. Chill.
@@VolksdeutscheSS hi Dear. I'm a teacher of Maths for over 49 years. I know confusing students with alternate tedious methods is unfair. So you need to be thankful for people like us that make the work load of students appreciative of us humble professions also. I normally do engage more able students to offer alternate math solutions.
@@VolksdeutscheSS Why negativity? I love to see everyone is engaging and offer alternative approach and having fun.
Brilliant ! I like the fact that the explanation is always easy with you, step by step, reminding the rules.
Glad you enjoyed it!
We can still simplify the problem in the following manner. Let x = a² and y = b². So that x - y = a² - b² = 9 ... (1). Again xy = a².b²=(ab) ² = 3² = 9 ...(2) Now using the algebraic equation (x + y )² = (x - y )² + 4xy , we get (x + y )² = 9² + 4. 9 = 117. So that (x + y ) = √117. Here negative sign is not considered as x and y are square terms of a and b . Replacing the values of a and b for x ad y, we get a² + b² = √117. Now (a + b) ² = (a² + b²) + 2ab = √117 + 2. 3 = √117+6. Therefore a + b = +-√ (√117+6), thus our result.
This was my answer too. Is it correct?
Thank you
simple and easy solution hats off
Very easy solution
Sir we can also do it by a+b=9/a-b then use(a-b) ²=(a+b) ² - 4ab and solve the quadratic
Nice bro
How will you solve this when we don't know a-b?
@@sabarishnarayanan4818 a-b=(a^2-b^2)/(a+b)
@@sanjayshinde5737 ya but we don't know a-b or a+b so it doesn't give us anything
@@sanjayshinde5737 nevermind, I got it
A shorter solution: Substitute x=a+b, y=a-b. We obtain from here that xy=9, x^2-y^2=12. We can remove y from this system to obtain a biquadratic equation for x
Dude how is that shorter you simpley asking the same question with more copmlex way
@@EfSaNe2531 In a way it is more straightforward, since we're interested in x=a+b, not a or b in itself. The other thing to note is the use of binomic formulas, the 3rd occurs obviously in the 1st eqn of the problem, the 1st and 2nd more indirectly as (a+b)^2-(a-b)^2=a^2+2ab+b^2-a^2+2ab-b^2=4ab, which connects with the 2nd eqn of the problem.
I.e. after recognizing the 3rd binomial equation and noticing that we're really interested in a+b, not a or b in itself, it makes sense to change variables to x=a+b, y=a-b and see how the second equation of the problem can be expressed in x and y, then eliminate y. That solution path always has the solution in its view.
The video solution has a lot of steps that only make sense later. E.g. why calculate a^4+b^4? Only when you already know where you're going next that makes sense. And even that is then first leading to a^2+b^2, another intermediate step where you have to know how that is going to help you.
@@Pengochan this guy just shows us the solution but doesnt teach us HOW to come up with it
@@jootpepet Yes, it's a kind of doing math like a stage magician: lots of handwaving (term transformations), then the solution comes out of nowhere.
Doing lots of tricks without really explaining how it is done. Of course each transformation in itself works out, but it isn't motivated, why it is done.
Of course sometimes one way to arrive at a solution is to play around with the equations, and in the end pick the way that works, but it's kind of stumbling around in a labyrinth, then presenting the way you found without ever mentioning all the dead ends.
@@Pengochan sir this is about practice. No-one can teach u shorter solutions in short time but u urself had to do practice so that those short tricks come in mind at right time. If u want explanation then it is " as question is asking to find a+b instead of a or b . Just rearrange both equation in form of a+b and a-b." This method will come in ur mind if u have done practice otherwise it won't come until unless u r too smart who don't need practice.
Shorter solution : i is the imaginary number
(a + ib)^2 = a^2-b^2+2iab
= 9+6i
Then |a + ib|^2 = sqrt( 81 +36 ) = 3*sqrt(13)
ie a^2 + b^2 = 3*sqrt(13)
a + b = + or - sqrt( a^2 + b^2 + 2ab)
= + or - sqrt(3*sqrt(13) + 6)
I recommend this maths problem . This guy is crazy applied a method that I have seen before in TH-cam .
th-cam.com/video/z2OyVIJznHw/w-d-xo.html
With Analisys Complex Number Solution.. 👍👍
6:18
a²+b² can't be negative, is it? So, we can write just 3√13, without ±
Square can be negative due to complex numbers wth a real and imaginary parts.
@@ichvvhovxugcvuhi3522 If a or b can be complex numbers, then at 8:00 still cannot be supposed that (a + b)^2 >= 0. If a and b are supposed to be real numbers, in the video he can reject the wrong negative solution at 6:00 instead of 8:00 as @gold user stated.
@@zsatmari a and b are not conjugates so the product/sum can be positive.
Solve simultaneously: b^2=-4.5+-1.5sqrt13
a^2=9/(-4.5+-1.5sqrt13)
@@ichvvhovxugcvuhi3522 but we're dealing with real numbers only.
@@chessandmathguy but the question didn't specify. Better to be inclusive of all possible answers real or complex. When the question specify other conditions then reject the respective value(s).
I love it!!! Very similar to "the answer is made by rotating a roulette N^((i*pi*e)^-i) times" Because exist traditional method solving this question
I followed the same approach, but a^2 + b^2 cannot be equal to -sqrt(117), since it is always positive.
You can go directly from a^4 - 2a^2b^2 + b^4 = 81 to a^4 + 2a^2b^2 + b^4 = 81 + 4a^2b^2. The LHS is (a^2 + b^2)2 and the RHS is 117. So instead of adding 18 twice you just add 36 once. And you don't need to consider a^2 + b^2 = - 3 sqrt(13). You keep that option around far longer than necessary.
consider the system (1)
a ^ 2 + b ^ 2 = k (where ^ 2 stands for squared)
a ^ 2 - b ^ 2 = 9
from which adding and subtracting the two equations member by member
a ^ 2 = (k + 9) / 2 and b ^ 2 = (k - 9) / 2 (2)
a ^ 2 * b ^ 2 = (ab) ^ 2 = 3 ^ 2 = 9,
applying the (2)
k ^ 2 - 81 = 36 k ^ 2 = 117 = 13 * 9 and k = ± 3 √13
from (a ^ 2 + b ^ 2)> 0, k = + 3√13 follows
a ^ 2 + b ^ 2 = (a + b) ^ 2 -2ab = + 3√13 and
(a + b) ^ 2 = 6 + 3√13 (the solution 6 - 3√13 must be discarded because it is negative in contrast to the first member of equality which is always positive)
a + b = ± √ (6 + 3√13)
I actually went through this approach
Actually, this is a very simple and trivial problem that can be solved directly by expressing unknown a (and b) from the second equation ab=3, and by substituting it to the first equation. In this obvious method, we get a quadratic equation for a^2 (and for b^2). Having received a^2 and b^2, it is very easy to find (a+b)^2 and solve the problem. Such a straightforward solution does not require any tricky actions (as it is done in this video). In my opinion, this problem does not correspond to the level of Math Olympiad, since it is very simple, and it can be done to the students in any regular (not math) class as a homework. BTW, since we are dealing with real numbers only (not complex), it’s absurd to use a minus sign in the expression (a^2+b^2) because it can't be negative.
I recommend this maths problem . This guy is crazy applied a method that I have seen before in TH-cam .
th-cam.com/video/z2OyVIJznHw/w-d-xo.html
That's what I've also noticed. This video complicates things
@@trexdantea9458 I think the host of the channel is trying to teach people substitution algebra. All his videos use this method. That's why it's used here even though there's a simpler way to do it.
In an examination this question carries 1 or 2 marks. This method would be too long. In my own view just make a or b the subject of the formula in the second eqn and substitute in the first eqn. Get quadratic equation. Solve the period
b=3/a
a^2-(3/a)^2=9
a^2-9/a^2=9
a^4-9a^2-9=0
a^2=(9+3sqrt13)/2. (a^2 is +ve -sign is Cancel)
Now putting the value of a^2 in Ist eqaution
Since a^2 - b^2 = 9 and ab = 3 then we can write a^2 - b^2 = ( a-b) (a+b) = 9 that means a + b = 9 / a - b if we substitute a = b/3 in that equation to becomes 3/b + b = 9 / (3/b - b) then by solving this equation we will get the following form : b^4 + 9b^2 - 9 = 0 now let's call b^2 = x then we get x^2 + 9x - 9 = 0 and by solving this equation we will get x = 0.908 and we know that x = b^2 that means b = 0.953 then a = 3, 1477 and a + b = 4,1 and that is the same answer. Thank you very much sir.
An aproximation of the solution as
sqrt(6 + 3 * sqrt(13)) is irrational.
That is basically what I did to solve the equation. I used the letter "u" instead of "x" and solved for a instead of b. I ended up with a^4 -9a^2 -9 and plugged in u for a^2, u^2 - 9u - 9 = 0. Solving gives the approximate values of u = 9.908327 and u = -0.953062. Substituting back for a^2 means a^2 = 9.908327 and a^2 = -0.953062. a = +/- 3.147750 and a = +/-0.953062 * i (which is rejected), so a = +/- 3.147750. If b = 3/a, then b = +/- 0.953062. (a + b) is approximately +/- 4.100812.
Super, Wafiq
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
I suppose it would be a lot easier if you use the concept a square-b.square= (a+b)(a-b)=9 and it gives a+b =9/a-b, given the ab =3, a+b =9/a-a/3, a+b=27/2a, giving a.square=81/2, giving a=9/root2 ,b now=3/a=3xroot2/9=root2/3; therefore.a+b= 9/root2+root2/3, a=b =29xroot2/3
Set t := a + b. Because (a-b)^2 = t^2 - 4ab = t^2 - 12, we have 81 = (a^2-b^2) = (a-b)^2(a+b)^2 = t^2(t^2-12), or equivalently t^4 - 12t^2 -81 = 0. Calculating the positive root of x^2 - 12x - 81 = 0 and taking ±sqrt of it, we obtain the values of t.
The most important thing is...We know that {ab = 3}
and{ a^2-b^2 = 9} means (a+b)(a-b) = 9.
and What we are finding is (a+b) NOT (a, b). better to ignore each of a, b.
At this point, the best approach is...
Just Put X = (a+b),Y = (a-b)
and... use Basics of math, below --> DONE.
[1]. (a+b)^2 = a^2+b^2 + 2ab
[2]. (a-b)^2 = a^2+b^2 - 2ab
Obviaously, [2] = [1] + 4ab ,and we know (ab=3).
[3]. Y^2 = X^2 + 12
[4]. XY=9
with [3] and [4], we get quadro of func(X), and find X^2, and find X.
(still skipping each of a , b).
a^2 - b^2 = 9
a.b = 3
So, a = 3/b and b = 3/a
a^2 - (3/a)^2 = 9
a^2 - (9/a^2) = 9 multiple by a^2
a^4 - 9 = 9a^2 (substitution)
a^4 + 9a^2 + 9 = 0,
Remember : ax^2+bx+c = 0
X1 = [-9 + root of 9^2 - 4x1(-9)] / 2
X1 = [-9 + root of 81 - (-36)] / 2
X1 = [-9 + root of 117] / 2
X1 = (-9 + 3 times of the squared root of 13) / 2
Or : (-9 + 10,82) / 2
1.82 / 2 = b = 0.91
And then,
If a = 3/b and b = 3/a
a = 3/0.91 = 3.29
So, "a" + "b" is equals to 4.2 (four second) or (four over two)
🥵 did it well..
Hi there and thanks for this nice system. Another shorter solution is to take X=(a+b)^2 and Y=(a-b)^2 .
We can easily get that X-Y=12 and XY=81. It’s then easy to find X and then a+b
Let a + b = c. By an express from this equality `a`(b) and squaring both sides we are getting a^2 - b^2 = c^2 - 2ac and b^2 - a^2 = c^2 - 2bc. Hereof c^2 - 9 = 2bc and c^2 + 9 = 2ac, from product of the last two equations are getting c^4 -12 c^2 - 81 = 0. By solved this biquadratic equation we are getting the same solution.
I have seen some of videos. How do you come up with a strategy for solving these equations? The connection from the start to the end is not obvious. What kind of intuition to you use?
you can just reject -ve at eq3, as a^2 + b^2 can't be neg for a,b in real numbers.
I recommend this maths problem . This guy is crazy applied a method that I have seen before in TH-cam .
th-cam.com/video/z2OyVIJznHw/w-d-xo.html
you had a^2+b^2= +-3sqrt(13), the negative can be rejected here as the LHS is non-negative
My attempt before video.
Let u=a+b, then we have
u² = a² + b² + 6 = 15 - 2b², and
(9/u)² = (a - b)²=a² + b² - 2ab = 3 - 2b²
Subtracting, both of them give
u² - 81/u² - 12 = 0.
⇒ u⁴ - 12u² - 81 =0
⇒ u² = 6 ± 3√13
⇒ u = ±√[ 6 ± 3√13].
Great video as always, found it even trickier than most of your other videos 😁
I am wondering: Could you solve a task like this one quickly if you were asked? I think it is anything but straight forward.
Thank you for your feedback! Cheers!
These kind of problems require slick moves, consistent practice, and continued exposure. No wonder I reveal all kind of question for my audience that require various techniques...
You are awesome, Tobias. Keep it up 😀
And you are COOL!@@PreMath
He can save a little time by dropping the solution -3(13)^(1/2) from equation (3) by noting that a^2+b^2 is >=0.
This one stretched me further than most of your other videos, so I will have to go through it again more slowly. I got part way without watching the video first, but I think remembering those identities is important for me.
8:48 the numerical value of + - sqrt (6+3 sqrt 13) is + and - 4.10609
Thank you for your feedback! Cheers!
You are awesome, Devon. Keep it up 😀
a*b=3 so b=3/a
Sub 3/a in a^2-9/a^2=9>>>let a^2 =x
X -9/X = 9
X^2 - 9X - 9 = 0. Solve for X which solves for a^2 which gives the value for a and b
Fulgurant résolution :
Aristarchus identity : (a^2+b^2)^2 = (2ab)^2 + (a^2-b^2)^2
Thus : (a+b)^2 = 2ab +(a^2+b^2)= 2ab + sqrt[(2ab)^2 + (a^2-b^2)^2].
Which gives immediately the two general solutions for a+b.
For particular solutions use : ab=3 and (a^2-b^2)=9 to get : (a+b)^2 = 6+3sqrt(13)
QED
Simple and standard way to solve this is:
subtracting the second equation multiplied by 3 from the first, we get:
a^2 - 3ab - b^2 = 0, This is homogeneous equation - sum of all powers is 2. And then b = k * a - is standard substitution for homogeneous equation.
a^2 - 3 a^2 k - a^2 * k^2 = 0; a 0;
k^2 + 3k - 1 = 0;
k = (-3 +- sqrt(13) / 2; and ab = 3 , hence a^2 * k = 3, hence k > 0, and the only k = (-3 + sqrt(13)) / 2.
and then from a^2 * k = 3, a = +- sqrt( (3/2) * (sqrt(13) + 3) ), and from b = k*a
b = +- sqrt( (3/2) * (sqrt(13) - 3) ).
Well done. I got the answer in almost the same way.
What are the practical applications of this?
consider the system (1)
a squared + b squared = k
a squared - b squared = 9
a ^ 2 or b ^ 2 = a squared or b squared
from which adding and subtracting the two equations member by member
a ^ 2 = (k + 9) / 2 ... b ^ 2 = (k - 9) / 2 ... (2)
a ^ 2 * b ^ 2 = (ab) 2 = 32 = 9, applying (2)
k ^ 2 - 81 = 36 ... k ^ 2 = 117 = 13 * 9 and k = ± 3 √13
therefore a ^ 2 + b ^ 2 = + 3√13 (sum of two squares necessarily positive)
a ^ 2 + b ^ 2 = (a + b) ^ 2 -2ab = 3√13 ... (a + b) ^ 2 = 2ab + 3√13.
(a + b) ^ 2 = 6 + 3√13
a + b = ± √ (6 + 3√13)
At 7:54 why are you rejecting the negative value? The problem as presented does not rule out a complex number solution.
I am trying to figure out how to do a simple math problem, my daughters teacher couldn't help. If I buy a stock for $30 dollars and it goes down to $28.75 and I end up losing $250. I am trying to find out how many shares I had in order to lose $250? I am trying to figure out what steps I have to do In order to know how many shares of a stock I need to buy In order to lose the same amount on different trades. Thanks for any responses.
There is a much more general, systematic, elegant, concise and faster way (which is quite surely the optimum), by use of the Aristarchus identity, whose modern version is the complex number quadrance (or modulus) fundamental theorem : Q(z)Q(Z) = Q(zZ)
Where z=a+ib, Z=A+iB, and Q(z) = a^2+b^2 is the quadrance of the complex number z.
Indeed in the special case z=Z it simplifies to : (a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2
From that starting point the general solution to the problem follows immediately :
(a+b)^2 = 2ab +(a^2+b^2)= 2ab + sqrt[(2ab)^2 + (a^2-b^2)^2]. (the negative sqrt is excluded by positivity of the LHS)
Which gives the two solutions for a+b. QED
I may have done an equivalent thing as you describe (but without really being sure if it's legitimate or not--though it does give the numerically correct answer). Basically, completing the square with complex numbers. Multiply eqn (2) by 2i and add to eqn(1). Then you get a^2 +2abi - b^2 = 9 +6i. The left hand side simplifies to (a+bi)^2. Then take the complex square root of the right hand side; a and b are the real and imaginary parts of the result.
@@bradwilliams7198 No! All what you've writen is perfectly correct, but unfortunately you are making a CIRCULAR reasoning, that you don't see yet, since you didn't finish your calculation.
Indeed, you get to the edge point of computing the square root of 9+6i. But how do you actually compute it?...
Obviously by looking for two real numbers a and b which satisfy the equation : (a+bi)^2 = 9+6i. And then what? Well you have little option else than developing this identity, which leads you to...your starting point?
Isn't it so?... 😉
That's why you made a CIRCULAR reasoning, hidden in you unfinished resolution.
Why is that hapening ? Because you only rewrite the problem in an other form, without actually decreasing it's "entropy". You didn't actually FACTORISE, even thow you have made a factorisation. But the one you're making, though correct, DOESN'T lift the problem to a lesser "entropic" level, which is only done by Aristarchus identity that I expose.
Such identities are corner stones. You canot avoid them. They are the keys of algebraic properties, like the very EXISTENCE of "complex numbers".
In such way that it's the lack of existence of a similar indentity in 3D, that prevent a "complex structure" to existe in 3D. And so on in 5D, 6D, 7D, 9D, etc. Only in 4D and 8D, such identities exist, similar to Aristarchus one in 2D.
In other words, though they may seem to be several ways to solve the problem. There is in fact only one! All correct ways will necessarely use in a way or another the Aristarchus identity, which is THE door keeper of 2D world.
Furthermore, I didn't actually make any use of complex number in my resolution. I simply apply directly Aristarchus indentity which gives almost immediately, with hardly any computing, the two opposit solutions :
(a+b)^2 = 6+3√13.
My mention of complex number was only to recall one of the fastest, concise and easy to remember computable way to derive Aristarchus identity. But such identity was discover thousands of years before the discovery of complex numbers by Tartaglia, Cardano, Ferrari, Bombelli,...in the XVI th century. The identity holds by itself already on pure Real numbers.
Nevertheless it opened the door, long before their discovery, to the very EXISTENCE of this for long hidden multiplicative structure on 2D vectors, that we call "complex numbers".
Thus to conclure this short glance in Mathematics History and hidden Arcanes, the very existence, or lack of existence, of such corner stones algebraic identities, tells us about the somehow MORPHOLOGY of spaces, in different dimensions.
In other words, it is highly non trivial nor innocent, to add a dimension... It may change a hell of things!
In 4D, though the QUATERNIONS exist, thanks to a similar Real algebraic 4D identity, COMMUTATIVITY nevertheless is LOST!
And furthermore in 8D, where the existence of OCTONIONS arises thanks to, again, the "miraculous" existence of a similar Real algebraic identity, it's even worse than with the QUATERNIONS, since not only COMMUTATIVITY is lost, but also the important arithmetic ASSOCIATIVITY. Which starts to bring OCTONIONS at the slipery edge of USEFUL numbers world!
@@Igdrazil Well, the easiest way to take the square root of a complex number (e.g. a +bi) is to convert to polar coordinates r = (a^2 + b^2)^0.5 ; theta = arctan(b/a). Then the square root will be (in polar coordinates) r^0.5, theta/2 (or theta/2 + pi for the second solution). Then convert back to cartesian coordinates.
So (9+6i)^0.5 = 117^0.25*(cos(0.5*arctan(6/9)) +i*sin(0.5*arctan(6/9)). Numerically, 3.14774... + i * 0.95306...
Write X = a+b, such that a-b=9/X, using (a+b)(a-b)=9.
Obviously (a+b)^2 - (a-b)^2 = 4ab, and thus X^2 - 81/X^2 = 12, such that (X^2)^2 - 12*X^2 - 81 = 0.
Solving this quadratic equation gives X^2 = 6 + sqrt(117), and thus a+b = X equal to sqrt(6 + sqrt(117)) and -sqrt(6 + sqrt(117)).
Great work 🙂👊
6:07 it can't be
There is a more simple solution: From handling Pythagorean triples we know, that a, b, c in an rectangular triangle may be represented by the expressions m^2 - n^2, 2mn and m^2 + n^2 for a and b, c respectively. Let x, y and z be the 3 sides of an rectagular triangle. So if a^2 - b^2 = 9 = x and ab = 3, y has to be 2ab = 6. Thus, z will be 3*sqrt(13). So (a+b)^2 = a^2 + b^2 + 2ab is equal to 3*sqrt(13) + 6 and ab = +/-sqrt(6 + 3*sqrt(13)).
Great and clear explanation Sir .
Thank you so much.
Thanks and welcome
Kind of a prety generic solution.
X = +/- sqrt(2V+sqrt(U^2+4V^2))
where U, V are used in place of 3 and 9.
It is also easily solvable by (a²-b²)²= (a+b)²(a-b) ²
Then (a²-b²)²=(a+b) ²{(a+b) ²-4ab}
Yes
Yes
Why did you reject the second case, it could have been done with complex numbers
At the end of Step 1 you had a^4 + b^4 =99. It seems like you should have been able to take the 4th root of both sides and ended up with a + b = 4√99.
But 4√99 does not equal +/-2√(6 + 3√13).
One more way is to take (a+bi)^2=a^2+2ab-b^2=9+6i and use square root formula for complex numbers to get an answer
i think i've found a better way to solve, we can use quadratic equations:
M = A-B
S = A+B
delta = b^2 - 4ac => M = sqrt(b^2 - 4ac) / |a| => in this case we divided the whole equation by a, so a = 1 and equation is:
P = A*B = 3 => x^2 - Sx + 3 = 0
and delta = S^2 - 12
so since a = 1 => M = sqrt(S^2 - 12)
(a+b) (a-b) = a^2 - b^2 = 9
S * M = 9
S * sqrt(S^2 - 12) = 9
sqrt(S^2 - 12) = 9 / S
S^2 - 12 = 9^2 / S^2 => S^2 = t
t - 12 = 81 / t
t^2 - 12t - 81 = 0
t = 6 + 3 * sqrt(13) => S^2 = 6 + 3*sqrt(13) => S = (A+B) = sqrt(6 + 3*sqrt(13))
~~t = 6 - 3 * sqrt(13) => since 3 * sqrt(13) > 6, so this is not acceptable~~
Your explanation is so “ step by step” and i actually follow the procedures !
Thank you for sharing ……Abe ( uk )
I am delighted that i got the right answer.
Thanks for video.Good luck sir!!!!!!!!!
You're welcome!
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
I just went for x=a² and y=b² and solved (1) x-y=9 and (2) xy=9 for x and y and simply added the two square roots.
My idea was different from this.
I had some idea
1)a,b>0 2)a,b0
a/b=(3+sqrt13)/2
b/a=2/(3+sqrt13)
sqrt a/sqrt b +sqrt b/sqrt a= (a+b)/sqrt 3
But my Renault for a,b>0 Is (sqrt 3)*{(5+sqrt13)/[(sqrt2)(sqrt(3+sqrt13)]
Its the samé on calculator, but not so nice .....
I wonder how to check such an answer?
a² - b² = 9 (1)
ab = 3 (2)
Lets divide (1) on (2)
a/b - b/a = 3 (3)
x - 1/x = 3 (4)
x² - 3x -1 = 0 standard square equation with solutions m
so
ab = 3 and a/b = m
a+b = √(3m) + √(3/m)
where m = (3 + √13)/2
Notes : can't use second root
(3 - √13)/2 since it's negative.
Alternatevely, a + b =
√(a² + 2ab + b²), so
a+b = √(3m + 6 + 3/m)
My solution below:
a + b = +/-sqrt[(a+b)^2] = +/-sqrt(a^2 + 2ab + b^2) = +/-sqrt(a^2 + 6 + b^2) = +/-sqrt[a^2 + 6 + (a^2 - 9)] = +/-sqrt(2a^2 - 3)
b = 3/a, so plugging that into equation 1 gives: a^2 - (3/a)^2 = 9
Rearranging gives a^2 - 9 - 9/a^2 = 0
Multiplying by a^2: (a^2)^2 - 9a^2 - 9 = 0
This is a quadratic equation in a^2, solving for a^2 gives: a^2 = [9 +/- 3sqrt(13)]/2
Plugging this into the equation found earlier:
a+b = +/-sqrt(2a^2 - 3) = +/-sqrt[9 + 3sqrt(13) - 3] = +/-sqrt[6 + 3sqrt(13)]
em....I think the question should mentioned that a and b is real number at the very beginning. If it state a and b is natural number instead, then obviously it is no solution
ab = 3 imply both a and b is odd number
(a+b)(a-b) = 9 which doesn't means sense as both a+b and a-b can only be even number, product of 2 even numbers can only be even number, so a+b is no solution
What about :
a^2-b^2-2ab = 9-6
-(a^2+b^2+2ab) = -3
-(a+b)^2 = -3
(a+b)^2 = 3
(a+b) = SQRT(3)
I don"t see where i miss something. Do you ?
Step 2. -(a^2+b^2+2ab) = -a^2-b^2-2ab. It does not = +a^2+b^2+2ab (I put + sign in front to emphasize that it is positive)
In other words, multiplying by -1 as you did in step 2 causes the a coefficient to be negative, where in step 1 it is positive.
Hope I could help! 😁😁
What is the software you are using can you please help?
What do a and b equal, separately?
I expressed a=3/b, substituted "a" in the first equitation with 3/b and solved quadratic equation.
How the hell is 6-3✓3 less than 0 Mr
2 equations and 2 unknowns. Just equate one of the equation to x. Then substitue this X equation to the second then simplify.
Professor;
if a²-b²=9 -> (a+b)(a-b)=9 -> a+b must be 9 b/c (a+b) >(a-b)
It is possible also if we take a=5 and b=4
Thanks for solving this. First, l want to say this solution is perfect but l used another algebraic formula for this question. I used a⁴+b⁴=((a+b)²-2(ab))²-2a²b². And when l watched this solution l saw it is better.👍
This types of maths which standard teaches in school
I learned, the result of sqr ist defined as positive, only the result of x-equations could be ±. How do You see this point?
That was a fair amount of work. Good explanation
if I wrote down: +- square root of (4.5 + square root of 29.25) + 3 : +-square root of (4.5 + square root of 29.25) does that count as the correct answer? The numerical value of this expression is the same as the value of the expression that the author of the video received, it's just that my record is longer (if I didn't have to write the square root in words all the time, it would be shorter). Sorry for my english, it's not my native language
I think this is shorter
from equation (2) ab = 3 then we get a2 b2 = 9
then substitute in equation (1)
a2 - b2 = a2 b2
a2 - b2 - a2 b2 =0
a2(1-b2) - b2 = 0 (3)
form equation (2) we get a=3/b or a2=9/b2 then substitute in equation (3)
(9/b2)(1-b2) -b2 =0
let b2 = y then we get
9(1-y)/y - y = 0 , multiply by y we get
9-9y-y2=0 (4)
y=(9 [+,-] Sqr(81 - (4 * 9 * -1) ))/(2*9)
y=(9 [+,-]Sqr(117))/ 18 we know that y must be positive and Sqr(117)>9 then
y=(9+Sqr(117))/18 = 1.100925213
b = [+,-] Sqr(y) =[+,-] Sqr(1.100925213) = [+,-] 1.0492
a = 3/b =[+,-] 3 /1.0492 = [+,-] 2.8592
a+b = [+,-] (2.8592 + 1.0492) = [+,-] 3.9084
The path you expose here, toward the correct solution, is good. Nevertheless a bit lengthy with furthermore some extra missed shortcuts like your lack of simplification with a2+b2=3sqrt13.
But remarquably, there is a much more general, systematic, concise and faster way (which is very probably the optimum), by use of the Aristarchus identity (which modern form is the complex number quadrance (or modulus) fundamental theorem) : Q(z)Q(Z) = Q(zZ)
Where z=a+ib, Z=A+iB, and Q(z) = a^2+b^2 is the quadrance of the complex number z.
Indeed in the special case z=Z it simplifies to : (a^2+b^2)^2 = (a^2-b^2)^2 + (2ab)^2
From that starting point the general solution to the problem follows immediately :
(a+b)^2 = (a^2+b^2) + 2ab = 2ab + sqrt[(2ab)^2 + (a^2-b^2)^2]. (the negative sqrt is excluded by positivity of the LHS)
Which gives the two solutions for a+b
QED
I recommend this maths problem . This guy is crazy applied a method that I have seen before in TH-cam .
th-cam.com/video/z2OyVIJznHw/w-d-xo.html
Is it possible to further simplify the final solutions?
The best way for you please trying solving this task using complex analysis method, then you Will finding imaginary and real component and then those shall be completed to solve
what if a+b is imaginary unit? question should suggest that a+b is real number
From two given equations, we have a^2=(9+-3rt13)/2.
Also from the two given equations we have (a+b)^2=2a^2-3, so we get the answer.
This problem is a little harder than it looks. My solution is fairly similar to professor's. First take (a-b)*(a+b)=a^2-b^2=9 (equation 3). Square this to get (a-b)^2 * (a+b)^2 = 81 (equation 3a). Next note that (a+b)^2-(a-b)^2= (a+b)^2-(a-b)^2=4ab=12 (equation 4). Rearrange this to (a-b)^2=(a+b)^2-12 (equation 4a). We can now get rid of the (a-b)^2 by inserting equation 4a into 3a. The result is [(a+b)^2-12]*(a+b)^2)=81. (equation 5). Now define u=(a+b)^2. Equation 5 then becomes (u-12)*u=81 (equation 5b) Or u^2-12u-81=0 (equation 5c). The solution to this quadratic is u=[12+-sqrt(468)]/2. Or u = 6+-3sqrt(13). Since u is a square and thus must be positive we reject the negative root 6-3sqrt(13), thus u=6+3sqrt(13). Since u = (a+b)^2, then a+b=+-sqrt[3sqrt(13)].
Very well done
Thank you for your feedback! Cheers!
You are awesome. Keep it up 😀
There is an unstated assumption that the solution must be in real numbers. If you allow complex numbers, then you cannot say that (a+b)^2 = 6-3(sqrt(13)) is a false statement.
This assumption should be included in the problem statement.
Curiously, the value for a is suspiciously close to pi. Not related in any way but close.
Why did we discount any imaginary solutions?
Very good lesson ..nice learn mathemathic program...suport
a square plus b square can only be a positive number. So one result can be removed in the third equation and no need to do that in Step 3.
eq-3 a square plus b square should be only positive. Because sum of square of two real numbers must be positve
The 3step,a^2+b^2=+3√13 or -3√13, we must delete -3√13 this time,
because we determin the answer 6-3√13>0,it is wrong
sqrt both sides
a-b=3
a-3=b
ab=3
(a)a-3=3
used a calculator and got sqrt6 for a
(sqrt6)^2-(sqrt6 -3)^2
6-6+9
6-6=0
0+9=9
a=sqrt6
b=sqrt6-3
I think this is right, but I'm about to go into highschool next year so I wouldn't be to surprised if it was wrong, is it?
I still do not understand why the second solution was rejected... there are lots of cases when the square of a number is a negative number. No problem.
Did the question specify a and b as real?
Thak you sir for this explanation . may i ask you about the name of books of Olympiad questions
Thanks for the visit, Vectoria
You're welcome!
We don't have a book. We just do constant search and planning. If I find any good book, I'd love to share with you.
Thanks for asking! Cheers!
You are awesome. Keep it up 😀
@@PreMath thak you ❣️
It is assumed that a and b are real numbers.
9²=(a²-b²)²=(a+b)²(a-b)²=(a²+b²+2ab)(a²+b²-2ab)=(a²+b²)²-4(ab)²=(a²+b²)²-4x3²
∴ (a²+b²)²=81+36=117. ∴a²+b²=√117=3√13. (a+b)²=a²+b²+2ab=3√13+6.
∴a+b=±√(6+3√13).
Yes you have used in fact a simplified form of the Aristarchus identity which modern version is the complex number quadrance theorem : Q(z)Q(Z) = Q(zZ)
You do not need to use the symbol '±' because it is a^2+b^2 > 0.
Okay, so I introduced new variables; p = a + b and m = a - b. p*m = (a + b)(a - b) = a^2 - b^2 = 9 and p^2 - m^2 = (a + b)^2 - (a - b)^2 = 4ab = 4 * 3 = 12. So I switched to a system with p and m, where I need to find p. p * m = 9 --> m = 9/p --> m^2 = 81/p^2. Substitute that in the second equation: p^2 - 81/p^2 = 12. Multiply by p^2 to get p^4 - 81 = 12 * p^2. Move everything to the same side and we get p^4 - 12 * p^2 - 81 = 0. This is a biquadratic equation, so we can solve for p^2 with the quadratic formula: p^2 = [-(-12) + - sqrt((-12)^2 - 4 * 1 * (-81))]/2 = [12 + - sqrt(468)]/2 = [12 + - 6*sqrt(13)]/2 = 6 + - 3*sqrt(13). 3*sqrt(13) > 6, so that cannot be an answer to p^2 (if we ignore complex answers). So p^2 = 6 + 3*sqrt(13). Which leads to p = a + b = + - sqrt(6 + 3*sqrt(13)).
Let X = (a+b)^2 = a^2 +2ab +b^2
Then summing this with the first equation leads to
2a^2 + 2ab = X + 9 hence 2a^2 = X + 3 [ab = 3]
Substraction gives
2b^2 + 2ab = X - 9 hence 2b^2 = X - 15
Let's multiply :
2a^2 x 2b^2 = (X+3)(X-15) = 4 x (ab)^2 = 36 which leads to
X^2 - 12X - 45 = 36
X^2 - 12X - 81 = 0
Delta is 12^2 + 4x81 = (2^2 x 2^2 x 3^2) + (2^2 x 3^2 x 3^2) = 6^2 x 13
Sqrt(Delta) = 6Sqrt(2^2+3^2) = 6Sqrt(13)
X is a square so > 0, only positive root is (12 + 6Sqrt(13)) / 2 = 6 + 3Sqrt(13)
finally a+b = Sqrt(6 + 3 Sqrt(13))
a=b/3 Substitut and solve quadratik equation and finish.
How do you arrive at plus or minus 3x square root of 13???????????????? This is what makes maths hard!
Please tell me how I can also solve these trickier questions as you solve it. LOVE FROM INDIA