Math Olympiad Question! Find the Missing Value in this System of Equations | Step-by-Step Tutorial
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- เผยแพร่เมื่อ 6 พ.ย. 2021
- Learn how to find the missing value in a system of equations by using FOIL and double distribution. Step-by-step tutorial by PreMath.com
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Absolutely brilliant!
Thank you Michael! Cheers!
You are awesome.
Keep rocking😀
@@PreMath
But how could x^2 + y^2 be -2
That's impossible
@@abdouraouf6863 It must be a
postive value not negative there is some thing wrong the two equations gives wrong values for x and y if we solved them in traditional way
@@abdouraouf6863 u r right
@@abdouraouf6863 x and y are complex number
It can be solved directly by the binomial expansion of (x+y)⁵
(x+y)⁵ = x⁵ + 5x⁴y + 10x³y² + 10y³x² + 5xy⁴ + y⁵
x⁵ + y⁵ = (x+y)⁵ - 5x⁴y - 10x³y² - 10y³x² - 5xy⁴
x⁵ + y⁵ = (x+y)⁵ -5xy(x³ +2x²y +2y²x +y³)
x⁵ + y⁵ = (x+y)⁵ -5xy((x+y)³ -x²y -y²x )
x⁵ + y⁵ = (x+y)⁵ -5xy((x+y)³ -xy(x+y))
x⁵ + y⁵ = 2⁵ -5*3(2³ -3*2)
x⁵ + y⁵ = 32 -15(8 - 6)
x⁵ + y⁵ = 2
I did this a totally different way. I used the first 2 equations and the substitution method to find values for x and y. Then just took them both to the 5th and added.
Me, too. It looked like a more direct path.
Well done, that's what I would do
This method is the best method .
This is NOTIN THE
This is the BEST METHOD not taught in the PRESENT SCHOOL TEACHING METHOD.
This highly VALUBLE IN
2nd degree equation solving etc.
We have to have good teachers with good QUALIFICATIONS teaching in schools FOR THE LOVE OF TEACHING SMALL GROWING STUDENTS .
TODAY MONEY IS THE " KING".
That seems to me much more obvious.
It’s important to notice that if the question mentioned anything related to real numbers like “system of equations over the real numbers “ there is no value , as the system implies that the sun of two squares is negative. Very good explanation !!
What is the domain of these values? I think they must be defined in the complex number set rather than the real numbers. Otherwise does not make sense like x^2 +y^2 = a negative value.
Yes, exactly. Either x = 1 + sqrt(2)*i and y = 1 - sqrt(2)*i, or vice versa.
Yes, not very satisfying to watch, probably not a premath problem
@@michaelmaggiano8540 I think that is the point: when you just go down the very simple path of turning 2 - y = 3 / y ... that turns into a quadratic equation. Solve that, power to 5, done. Of course, that only works when you deal with the complex numbers. The above works without that, but of course, is insanely more complicated.
Im proud of all y'all for asking such a question.
Don't bow to to false gods. Well done.
If either x or y is a complex number how can x times y be 3? Shouldn't x times y be complex also?
I went a slightly different route -- I used Newton' binomium. You expand (x+y)^5, then (x+y)^3, then you are left with a bunch of terms you can express in terms of (x+y) and xy. Fill in the values et voilá, it solves itself.
Great approach
You are awesome Ben.
Keep rocking😀
I did this as well. The video presents a solution the requires insight.. how would mere mortals come up with the FOIL to begin with? Bang out the binomial of the fifth power, collect the terms with the same constant, factor out the xy or (xy)^2, do the same with the sum of cubics, and done.
Details for this :
(x+y)^5 = x^5 + 5x^4y + 10x^3y^2 + 10x^2y^3 + 5xy^4 + y^5 with Newton's formula.
So x^5 + y^5 = (x+y)^5 - 5xy(x^3 + 2x^2y + 2xy^2 + y^3).
The only annoying part is the last factor, but as suggested we are close to (x+y)^3 = x^3 + 3x^2y + 3xy^2 + y^3.
Indeed x^5 + y^5 = (x+y)^5 - 5xy(x^3 + 3x^2y + 3xy^2 + y^3 - x^2y - xy^2)
So x^5 + y^5 = (x+y)^5 - 5xy((x+y)^3 - xy(x+y)) = 2^5 - 5*3*(2^3 - 3.2) = 32 - 15*2 = 2.
Same method here. First idea was solving the implied quadratic, but I didn't want to bother with calculating the fifth powers of the complex solutions, and moving to (x+y)^5 is sufficiently natural.
@@Herfuage
흠..뉴턴공식을 이용해서 풀어도 되겠군요? 아주 좋습니다.감사합니다^^
I understand that your solutions for x,y are not restricted in R but are extended in C as well.
Otherwise, you should have stopped when you had found that x^2+y^2=-4 and conclude that there is no solution in R.
Am I correct? Or do I lose anything?
You're correct, I suppose...
Exactly; x and y have nonzero imaginary parts. The interesting fact: x^5+y^5 is real again; another example "how really existential" complex numbers are. Too bad we call it still "imaginary" and scare little children :-) from having a bias-free look into the concept.
@@WolfgangFeist In a way, they still are "imaginary". I've never taken a measurement of a distance, time, weight, etc. and come up with an imaginary result :)
Only when you get into advanced physics such as electrical phenomena does the imaginary component become "real".
@@ovalteen4404 well, in some quantum processes we do measure "imaginary parts" - of course at the very end transformed and transferred to a real "classical" instrument. But, you could say, the "real parts" of a quantum wave function or not any more 'real' then the imaginary.
By the way: It's even true if you use electrical complex calculation for e.g. an alternating current. If you have a capacity in the circuit, the phase shifted current ("blind-current") isn't any less "real" than the real part. :-) and we can certainly measure it using a scope. I'm just playing with names here; in complex number theory we just use "real" and 'imaginary' to have a clear notation of which cartesian corordinate of the number I want to point to. And then there is that philosophical meaning of "real" --- intuitively we all pretend to know what it is. But do we?
After all that crazy math, I got an answer of 2. I got x = 1±isqrt(2) and y = the conjugate of x. Then I used the binomial theorem for (a+b)^5 and (a-b)^5 and got the sum equal to 2a^5+20a³b²+10ab^4.
Excellent JSS
Thank you! Cheers!
Keep rocking😀
That was my approach as well. As soon as I saw that the real part was 1 I suspected that the imaginary parts would cancel since powers of complex numbers looks somewhat like rotations.
Is everything ok with x^2+y^2=-2?
How can the sum of 2 positive numbers can be negative ?
I wonder if you can comment on x^2 + y^2 = -2. The sum of 2 square being a negative number.
Whether you think this is a contradiction? Or what kind of complex number (x and y) we are dealing with?
Beautifully done! Merci (meaning: thank you)
love it a lot, thank you very much
Although I did not know how to answer this I learned a lot watching it thank you
Doing it the regular way, computing the complex values of x and y and then evaluating x^5+y^5 seems much easier to me.
That's what I did. Much more conventional and rational.
I did this, but I don’t think I would’ve gotten the exact value from computing sqrt(243)*2*sin(5*arctan(sqrt(2))).
@@chixenlegjo Hi!
I don't see why you're getting an arctan in there, but here's how I got the exact solution:
Substituting the second equation into the first one, I get
x+3/x = 2
multiply with x on both sides
x^2 - 2x + 3 = 0
so x = 1 +- sqrt(-2) = 1 +- sqrt(2) * i
Then using equation 1 to get y, we get
y = 2-x = 2 - (1 +- sqrt(2) * i) = 1 -+ sqrt(2) * i
So x and y are complex conjugates of each other, and we can simply continue with
x = 1 + sqrt(2) * i
y = 1 - sqrt(2) * i
Swapping x and y doesn't make a difference here.
Now we square x:
x^2 = (1 + sqrt(2) * i)^2
= 1 + 2 * sqrt(2) * i + 2 * i^2
So with i^2 = -1 we get
x^2 = -1 + 2 * sqrt(2) * i
And square it again:
x^4 = (x^2)^2 = (-1 + 2 * sqrt(2) * i)^2
= 1 - 4 * sqrt(2) * i + 8 * i^2
x^4 = -7 -4 * sqrt(2) * i
Now multiply with x to get
x^4 * x = x^5
= (-7 -4 * sqrt(2) * i) * (1 + sqrt(2) * i)
= -7 -7 * sqrt(2) * i -4 * sqrt(2) * i - 8 * i^2
So
x^5 = 1 -11 * sqrt(2) * i
Doing the same with y (or simply thinking about it geometrically), we get
y^5 = 1 +11 * sqrt(2) * i
So x^5+y^5 = 2
:)
I substituted y=2-x into xy = 3. I had to complete the square and got (x-1)^2 + 2 = 0. Therefore x=sqrt(2)*i + 1. Substituting back into the original equation you get y= -sqrt(2)*i + 1. Then I used Pascal’s triangle to know that x^5/y^5 would have coefficients 1 5 10 10 5 1. When you substitute your x and y values into this equation all the odd numbered exponents cancel out and you’re left with 10*(sqrt(2)*i)^4 + 20*(sqrt(2)*i)^2 + 2 = 40 - 40 + 2 = 2
I liked how you solved it in the video. It was really clever. Someone would need to be very skilled in mathematics to look at that problem and solve it the way you did.
Excellent!
I've solved for y (=2-x), and then quadratic roots for x (of x²-2x+3=0), x=1±i√2 and y=1∓i√2, used Pascals coefficients for power 5, 2nd 4th and 6th terms cancel out = no more i's, and got the same answer: x^5 + y^5 = 2.
How did you, how should we/I, intuitively(?) split to ^2 and ^3, etc.?
This is the most common mistake I've did. Don't solve for x and y separately. Instead, manipulate and play around with the polynomial to find. Let's start with a simple case: If x+y=2 and xy=3, what is x^2 + y^2? You can rewrite x^2 + y^2 as (x+y)^2 - 2xy, of which you can substitute both x+y and xy because it was given. They are getting easier if you know this trick.😀
@@randaya5854
I see - I should have "seen" x^2+y^2 as part of (x+y)^2 and then "look for" the "missing" xy(2).
What I want to know is how you typed the upside down version of (±). Is that some code with the ampersand?
Awesome Oren
Thank you! Cheers!
Keep rocking😀
@@XJWill1 Thanks...I often use alt codes. Hold the alt button on the keyboard and type a 4 digit code. Alt+0177 is the code for plus/minus.
Excellent video
After a lot of correcting algebra mistakes I got the answer using basically the same method he used. This is one of a big family of problems he's presented where a key (or the key) is to use (x+y)^n = x^n+y^n+... .
I also tried just solving the two given equations for x and y which leads to a quadratic that has complex solutions x=1+sqrt(2)i, y=1-sqrt(2)i, and the second solution with the +- signs reversed. I believe by symmetry they both give the same answer for x^5+y^5. After some complex arithmetic you get the same solution to the problem as by the first method. Both seem to be comparable in their level of tediousness.
I see that others have come up with this too but when you solve it on your own you want to take credit for that. (smile emoji)
by substitution from ist two equations , we have
x^2 =2x-3
x^3 = 2x^2-3x = x-6
and x^5= -11x+12
Same follows for variable y
y^5= -11y+12
adding both we get
x^5 + y^5 = 24-22=2
Note the recurrence relation
x^n + y^n = (x+y)(x^(n-1) + y^(n-1)) - (xy)(x^(n-2) + y^(n-2))
Define f(n) = x^n + y^n and a = x+y and b = -xy, and rewrite
f(n) = a f(n-1) + b f(n-2)
In this example, f(1) = a = x+y = 2 and b = -xy = -3. Also
f(2) = x^2 + y^2 = (x+y)^2 - 2xy = 2^2 - 6 = -2
Thus the sequence a=2, b=-3 becomes
f(1) = 2
f(2) = -2
f(3) = -10
f(4) = -14
f(5) = 2
Or in terms of the original problem
x^5 + y^5 = 2.
In general, every second degree recurrence relation can be summarized with a quadratic equation. Let x and y be the positive and negative solutions x+ and x- to a quadratic. (I can't do subscripts here, as I'd prefer.) Then
(x - x+)(x - x-) = x^2 - (x+ + x-)x + (x+x-) = x^2 - Sx + P = 0
x+/- = (x+ + x-)/2 +/- Sqrt ( (x+ + x-)^2/4 - (x+x-) ) = S/2 +/- D,
where sum S = x+ + x-, product P = x+x- and discriminant D^2 = (x+ + x-)^2/4 - (x+x-) = S^2/4 - P. Then
(x+/-)^2 = S^2/4 + D^2 +/- SD
The general recurrence relations are
x+^n +/- x-^n = (x+ + x-)(x+^(n-1) +/- x-^(n-1)) - (x+x-)(x+^(n-2) +/- x-^(n-2))
or
(x+/-)^n = (x+ + x-)(x+/-)^(n-1) / 2 +/- (x+x-)(x+/-)^(n-2) / 2
= f(n)/2 +/- D g(n)
where f(1) = x+ + x- = S, f(2) = (x+ + x-)^2 - 2(x+x-) = S^2 - 2P, g(1) = 1 and g(2) = x+ + x- = S. Then
f(n) = x+^n + x-^n
2D g(n) = x+^n - x-^n
In the given example,
S = 2
P = 3
D^2 = S^2/4 - P = -2
D = i Sqrt(2) (complex)
f(1) = S = 2
f(2) = S^2 - 2P = -2
g(1) = 1
g(2) = S = 2
a = S = 2
b = -P = -3
The characteristic quadratic
x^2 - Sx + P = 0
x^2 - 2x + 3 = 0
The solution
x,y = S/2 +/- D = 1 +/- i Sqrt(2)
x^n,y^n = f(n)/2 +/- D g(n)
f(n) = x^n + y^n
2D g(n) = x^n - y^n
I already gave the first five f(n) sums. Here are the first five g(n) differences divided by 2D, with the same a=2, b=-3.
g(1) = 1
g(2) = 2
g(3) = 1
g(4) = -4
g(5) = -11
In the case x+ = Phi = (1+ Sqrt(5)) / 2 and x- = - 1/Phi = (1- Sqrt(5)) / 2, S = -P = a = b = 1 and 2D = Sqrt(5).
x^2 - x - 1 = 0
f(n) = L(n) = the Lucas sequence (L(1) = 1, L(2) = 3)
g(n) = F(n) = the Fibonnaci sequence (F(1) = F(2) = 1)
These musings can be generalized to more than two variables (x and y or x+ and x-) with the Girard-Newton identities.
I do not know if it can be considered as solution if no pair (x,y) exists.
I have question:
What is in integer 4*x, if 2*x=3?
Will 6 be correct answer? Or no integer x exists?
awesome just loved it
Thank you
How can x ^2 plus y^2 be a negative number ?
OK, but please tell me, how could the sum of the squares of two numbers be equal to a negative number, in this case, -2? Unless one or both of them are imaginary numbers.
Thank you for teaching and using the phrase "double distribution" instead of FOIL. This is the way I like to teach my students.
I'm trying to understand your treatment of how to find x^3 +y^3. Is there a formula for factoring (x+y)^3 ? If so please explain to me its format.
I don't see how you finally get the extra (x+y) at the end and why is it multiplied to the x^3+y^3? I really would appreciate some clarification here.
Pascal’s triangle
(x+y)^3=x^3+3(x^2)y+3x(y^2)+y^3=x^3+y^3+3xy(x+y)
What are the values of x and y?
The following seems simpler: From the first two equations get x^2+y^2 = -2, squaring that gives us 4 in the right hand side of the equation.
Subtracting 2(xy)^2, or 2*3^2 gives us that x^4+y^4 is -14. The use x^5+y^5 = (x+y)(x^4-x^3y+x^2y^2-xy^3+y^4) = 2(-14-xy(x^2+y^2)+x^2y^2)) = 2(-14-3(-2)+9)=2.
(x+y) is a factor of x^n + y^n for any odd positive integer n, such as n=5 here.
Given, x + y = 2 and xy = 3
So, x^2 + y^2 = (x+y)^2 - 2(xy) = 2^2 - 2(3) = 4 - 6 = -2
So, x^4 + y^4 = (x^2+y^2)^2 - 2(xy)^2 = (-2)^2 - 2(3)^2 = 4 - 18 = -14
Since (x+y) is a factor of x^n+y^n for any odd positive n,
So, x^5 + y^5 = (x + y) ( [ x^4 + y^4 ] - xy [ x^2 + y^2 ] + (xy)^2 ) = 2 ([ -14 ] - 3 [-2] + (3)^2 ) = 2 (-14 + 6 + 9) = 2x1 = 2 . Simple. Right ?
I do not understand this question. When graphing the first two equations their lines do not intersect. So there are no good values for X or Y. So how can the untrue equations be used to calculate X^5 and Y^5 ? Can you please explain.
x and y are imaginary in these equations
x = 1±sqrt(-2). Due to to the inability to take the root of a negative number, this result is an "imaginary" solution for x. You will then get y = the conjugate of x. Then, amazingly, when you plug through the formula for x^5+y^5, the imaginary parts go away and you are left with a real solution.
how can X^2 + Y^2 be a negative number (-2) ?
Good shape of solve. Using rules.
Excellent
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Keep rocking😀
x & y use be complex nos for this to work.
The part of this equation is that x^2 + y^2 = -2. I would say that is not possible equation because every ingredient of this formula is raised to power 2 so it means that the sum of this ingridients cannot be below 0..?
You are brilliant sir
The problem was excellent
Sir.. I found
X=1+i.sqrt2
Y=1-isqrt2
Substituded values and got same answer but it was quite
Deficult
I let two ads play out to their ends so that you get paid and make more videos.
Bravo, your solution. Thank you for tutorial.
Wow, thank you dear
You are awesome as usual.
Keep rocking😀
Fun fact: x and y is complex number, with x = 1 + sqrt(2)i and y = 1 - sqrt(2)i
하하하하 맞습니다.동감합니다^^ 복소수라는 함정이~~ㅎㅎㅎ
If y=2-x then xy=3 becomes x^2-2x+3=0. We have that x=+/- sqrt(2)*i + 1, then y = -/+ sqrt(2)*i+1. Now we have that (sqrt(2)*i+1)^5 + (-sqrt(2)*i+1)^5 = 2, by Newton's binomial theorem.
Thanks
Welcome DD
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Talk about doing this the hard way. I solved the top equation for y (which is: y=2-x). Then substitute your 2-x into the 2nd equation to yield the quadratic: -x^2+2x-3=0. Solve for x: x=1+sqrt(2)i or 1-sqrt(2)i. Now that you have x's, solve for y's: y=1-sqrt(2)i or 1+sqrt(2)i. Insert x's and y's into x^5+y^5 and you get two answers that are the same: 2.
I solved for y in the second equation of the system and substituted for x in the first equation to find the solutions for x and y. I did up with complex numbers, so I used the binomial theorem in the last equation when solving it out, and I ended up getting the same answer of 2.
nice solution sir thank
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x^2 + y^2 = -2 therefore, x and y are on the circle, which radius is (square root of 2)*i. Another words, this circle has a range from - (square root of 2)*i to + (square root of 2)*i.
thus, x =(square root of 2)*i*cosA and y = (square root of 2)*i*sinA
and x.y =(square root of 2)^2*(i)^2*cosA*sinA = -2*(cosA*sinA).
Maximum of cosA*sinA is 1/2, when cos2A=0, or A is 45 degrees, or 225 degrees...
And Maximum of -2*(cosA*sinA) is 1 (when A is 135 degrees, or 315 degrees).
Therefore in this case, any |x.y| > 1 is invalid.
x.y = 3, No solution.
Also, x^2 + y^2 = -2, meaning, either x is an imaginary number, or y is an imaginary number, or both are imaginary numbers.
Then how come x^5 + y^5 is real number?
How is x^2 + y^2 = -2 possible??? Sum of two positive quantities is giving a negative quantity???
None of it seems possible. How is x^5 + y^5 = x + y? They can’t both be 1 or else xy = 1, not 3. America explain!
You're dealing with complex numbers here, ones that aren't also real numbers. Also even if x or y were real, squaring could technically give a 0 quantity as well as a positive one.
Wow sir! You have an amazing mind! Love from Pakistan 🌹🌹🌹🌹
How it is possible that x^2+y^2 be equal to the negative number?????
Nice problem and nice working
The answer is two , I took the hard way and expanded x+y=2 to the fifth power , solving along the way and at the end I'm back to square one with two as the answer
Excellent Vegeta
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Lmao
@@user-lh5hl4sv8z ikr
Muito boa as questões!!!
I tried using substitution and got lost. But this was fun to follow. I got to learn something new.
Thanks! Really a well done procedure, but I have a question: If "x + y = 2" (given equation), neither x, nor y can not be an irrational number. So how is it possible, that later (in an intermediate step) x^2 + y^2, a sum of 2 square numbers equals -2?
I spotted this, appears there are imaginary numbers involved here which doesn't make sense with the other equations
@@mikeo9863 Yes, that's correct! I've thought as well about involving i^2 = -1 into this problm, but e.g. if you take for y = 1 + i and y = 1 - i, of course it will sum up to 2, but you'll get for (1 + i)^2 = 1 + 2i - 1 = 2i and for (1 - i)^2 = 1 -2i - 1 = -2i, so that this sum will be 0, which also doesn't make sense, as it has to be -2 due to the shown procedure.
@@mikeo9863 Only with a combination for x, y = 1 +/- i sqrt(2) it might work, but what about the odd exponents in the other intermediate steps? - an interesting problem
i solved the first two equations together and got x= 1+sqrt(2) * i and y = 1-sqrt(2) * i . or vise versa. then substituted that in x^5 + y^5 and got 2 as well
Solving the first two equations for (x, y) = (1+i*sqr(2), 1-i*sqr(2)) quickly shows that x5+y5=2
how in the world is x square + y square equalss -2 . how can it be negative
It would be better to give the conditions on the variables at the beginning. This is something students need to develop early in their education.
The 2 different methods I used:
1) Simultaneous eq + quadratic formula to solve for x, y.
square x, y
square again to get x^4, x^4
multiply by x, y to get to x^5, y^5
add
2) Binomial expansion for (x + y)^5 = 2^5
simplify using (xy)=3 and (xy)^2=3^2 -> (x^2)*(y^2)=9
apply this method again for (x + y)^3 =2^3
use answer to substitute back into original
solver for x^5 + y^5
Great but what are x and y?
How that sum of squares is negative ?
I solved it the same way, but how is it possible that the sum of two positive numbers be negative???? (x^2+y^2)=-2 how this is possible???
4:08 sum of squares is negative?
Cool but a bit cumbersome. From x^2 + y^2 = -2 you see that the answer needs to be complex. Because of the xy symmetry, x and y are conjugate,. Their real part needs to be 1 (from x + y = 2) and their magnitude sqrt(3) (from (xy = 3) and thus the imaginary part +- sqrt(2), So we have x = 1 +i*sqrt(2), y = 1 - i*sqrt(2).
ALTERNATE METHOD
Fast forwarding to
Solving for x³+y³
We can expand it as
(x+y)(x²+y²-xy)
Substituting eq 1 and the given terms we get
(2)(-2-3)
(2)(-5)
x³+y³=-10 (equation 2)
On peut calculer x et y, on trouve des nombres complexes conjugués.
Sous leur forme exponentielle il est facile de les élever à la puissance 5
My solution was something similar
(x + y)^3 = x^3 + y^3 + 3xy(x + y) = 8 =>[simple calculation}=> x^3 + y^3 = -10
(x + y)^5 = x^5 + y^5 + 5xy(x^3 + y^3) + 10(xy)^2*(x +y) = 32 =>[another simple calculation]=> x^5 +y^5 =2.
Wouldn't it be easier to use two equations for solving X and y and then just bring each to the power of 5?
the system of equations (x + y) = 2 and (xy) = 3 has no (real) solution. Which is why other commenters talked about no solution existing in the real space R, but the solution does exist in complex space C. See the comment by Oren Ben-Dov for an explanation of how to find the roots. With the solution being the ordered pair (1±i√2, -1±i√2), then you have to brute force your way through foiling both of those messy irrational complex numbers four times without losing track of numerous ±n√2i and since I am too lazy to try it out by hand I don't even know if they will all cancel out easily.
@@Lawfair They have to cancel out easily otherwise 2 could never be the final answer.
This is straight but harder to solve.
Great
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Good solution.
Perhaps no real solution for X and Y.
Try to find different solution technique.
Pls help me for it Pascal's triangle coefficient method
I used the technique of finding x and y from the 1st 2 equations [x,y = 1 +/- i*sqrt(2)], then putting these values in the 3rd expression. Not sure if it's any quicker, but it would be if we needed to find other combinations of x and y as well.
@@andrewrao634 thank you
Dear Suresh, I will upload a video on Pascal's Triangle very soon.
Thank you! Cheers!
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I have recently started watching as I find them very interesting. I read all the comments posted as of the time of my post.
My question is for Step 1. I do not understand why this (x^2 + y^2) (x^3 + y^3) was written. Understand FOIL.
What am I missing?
The reason you begin by multiplying (x^2 + y^2) with (x^3 + y^3) is that you can create the (x^5 + y^5) term.
You want to multiply (x^2 + y^2) with (x^3 + y^3) in order to obtain a result with the x^5 and y^5 terms and other terms that you hope you can derive their value. And you also hope to be able to derive the value of (x^2 + y^2) and (x^3 + y^3).
One doesn't have to start that way. It's a choice, one that provides x^5 + y^5.
How can x2 + y2 be -2? Since the square of any number must be positive.
No, the square of any real number must be positive.
Yeah Sum of 2 square numbers. How can it be negative? I got the answer 2 too via a long method but still I find the above aspect odd
I am in the same boat
Something is not correct.
Complex numbers. i is defined as sqrt(-1)
Make x the subject in equation (1)
This will give you x=2-y
Put x=2-y into equation (2)
This will give you (2-y)*y=3
Expand to get 2y-y^2=3
Rearrange to get a standard quadratic equation y^2-2y+3=0
Solve and get y=1
Put y=1 into equation (1) to get x=1
Now put x=1 and y=1 into equation (3)
1^5+1^5 => 1+1=2.
Simple!
what is foil? what does it mean?
Foil => First Out In Last. It simply means double distribution.
Cheers Luffiz!
Keep rocking😀
@@PreMath thanks
Niceee. I did a hard way but come up with the same answer.
Все хорошо. Но x2+y2 не может быть отрицательно, разве нет?
If x^2 + y^2 = -2 then either x or y has to be an imaginary number, right?
They both have to be complex numbers with an imaginary term that cancels out when added. They cannot be both just imaginary as their sum would not be real.
What is the value of "x" and what is the value of "y" ?
Buen vídeo
Determine x and y
I figured I'd need a 5th power, so I wrote out (x+y)^5 and saw by inspection it was (x^5+y^5) + 5(xy)(xxx+xxy+xyy+xxx). The last bit becomes (x+3)^3 - (xxy+xyy), and the last bit of that is (xy)(x+y).
That is the most intuitive way to do this problem.
Practice ,the actual process is all logical,but I’m guessing practice to get any flow to answer
you should specify the domain. i.e in GF(5) the solution is the same but one should not do this magic at all only raise the first equation to the power 5.
very well done bro, thanks for sharing this math olympiad question
Thanks and welcome my dear friend
Cheers!
Keep rocking😀
How x2 + y2 can be negative ?
Complex world.
x^2+y^2=-2 NFW in R.
How is it possible to have a positive result with the addition of 2 squared numbers (a squared number is inevitably positive too)🤔
It needs to get into complex numbers, imaginary numbers. When you square i, you get -1. :)
This is unnecessary to the question at hand -- we don't have to find any actual values for x and y. But yes, their values would require i to be part of the solution.
I found out the value of x^2+y^2 as (-2)..and then compared the values of the variables in this equation, and the first equation x+y=2.
Solving the 2 equations, I got the value of both x and y, as 1..
x^5+y^5= 1^5+1^5=2
Sir, can you please tell me if my method is correct?
Surely x=y=1 does not fulfil the second equation. x and y are complex numbers, conjugate to each other if you want to go the brute force way.
Thank you so much for the correction, sir :)
#foil #foilmethod #binomial #polynomial
Newton Method. x + y = 2 and x*y=3, x, y are solution of quadratic equation X^2-2X+3=0. Then Newton polynomial is S(n+2)=2*S(n+1)-3S(n) with P(0)=2 [x^0+y^0=2] and S(1)=2
So x^2+y^2=S(2)=2*2 -3*2=-2. x^3+y^3=S(3)=2*(-2)-3*2=-10 x^4+y^4=S(4)=2*(-10)-3*(-2)=-14 and x^5+y^5=S(5)=2*(-14)-3*(-10)=2
Please use some time Newton`s Method.
Has anybody solved for x and y? I need to know these crazy numbers that produce such irregular results.
1 +- sqrt[2]i are the values for x and y (either order works). They're complex numbers.
The two first equations do not have real solutions. The complex solutions are x = 1 +/- sqrt(2)i and for each case y is the complex conjugate.
x^5 = 1-11sqrt(2)i and y^5 = 1+11sqrt(2)i (or vice versa). In any case, x^5 + y^5 = 2 as the imaginary parts cancel.
Yeah simultaneous equations.
x = 2 - y
(2 - y) * y = 3...... 2 = (3 / y ) + y
y = 1, so x + y = 2, so x = 1, so x^5 + y^5 = 2
x1=1+Sqrt2*i, y1=1-sqrt2*i; and x2=1-sqrt2*i, y2 =1+sqrt2*i
so, x^2 + y^2 =-2, and xy=3 satisfied !! (Please be noticed that: i^2=-1)
How X^2+Y^2=-2?
It is not specified in the statement of the exercise, no more in explanation, but S ∉ ℝ. In fact, x²+y² = -2 have no solution in ℝ. So S ∈ ℂ. In this case x=1+i√2 and y=1-i√2
It is important said that x,y are part of complex set, because they have solution find real values only for the operation itself. Something is wrong in the common sense when you figure out x2 + y2= -2, I could fell that something is wrong since x2+y2>=0, x,y E Real set.
my solution is
let t(n)=x^n +y^n
And I found out t(n)=t(n-1)*t（1）-xy* t（n-2）
to use this sequence, I need first find t(1) and t(2) which is simple
Then follow the sequence and find t(5)
c'est faut dans R(ensemble des reel)
From eq.I and II there is no solution with Real Number. If you draw graphs of these equations, they will not be bisected.
If x and y are real numbers, then the x⁵ + y⁵ has no solution.
@@diedoktor Which means the answer is UNREAL after going through all the trouble of using REAL algebra to come up with a solution.