This is the most complicated way to solve this problem. The simple way is to use equation 2 to to get y=(35/x) and substitute it into equation 1. With a little restructuring, you get a quadratic polynomial and use the quadratic equation to solve for x. Once you have the values of x, it is easy to get the values of y.
Just do it by inspection; no Quadratic Equation or even square roots to mess with. 1) Factor out x² - y² = 24. (x + y)(x - y) = 24 2) First, find the solution for only positive real integers. List the possible factors for 24 (in order). Note that since 24 > 0, x > y. (x + y) 6 8 12 24 (x - y) 4 3 2 1 i.e., [(x +y),(x-y)] = (6,4), (8,3), (12,2), and (24,1). 3) List the values of 2x & 2y for these ordered pairs. (x + y) + (x - y) = 2x (x + y) - (x - y) = 2y 2x 10 11 14 25 2y 2 5 10 23 4) List the values of 4xy = (2x)(2y). 4xy 20 55 140 575 5) The third value (140) yields the given initial condition. 4xy = 140 xy = 35 6) The third pair of factors is the solution for positive real integers. (x + y) = 12 (x - y) = 2 x = 7 y = 5 7) Negative (real) integers provide another branch. (x + y) = - 12 (x - y) = - 2 x = - 7 y = - 5 or (x,y) = (7,5) or (- 7,- 5) and (x + y) = ± 12 8) For complex solutions, note that 24 and 35 are explicitly real, so complex values for x and y can only be explicitly imaginary, i.e., x = i u y = i v 9) Factor out the original expression in terms of u and v. x² - y² = - u² + v² = 24 or (- u + v)(u + v) = 24 10) Use the method above used for real numbers to arrive at the solution for u and v. (|u|,|v|) = (5,7) 11) To find the quadrant of the imaginary solution, recall the condition xy = 35 The only way the product of two imaginary numbers can yield a positive value (i.e., 35) is if their signs are complementary. In other words, (5i)(- 7i) = + 35 and (- 5i)(7i) = +35 Thus, the imaginary solution is (x,y) = (5i,- 7i) or (- 5i,7i) and (x + y) = ± 2i
@zachowland Well, you're entitled to your own opinion. I solved the problem in a couple minutes using this method (without watching the video first), so I'm not sure I accept that my method is "infinitely" more complicated. In _my_ opinion, my method is simpler. 😉
From x^2 - y^2 = 24. Divide it by xy = 35. (x/y) - (y/x) = 24/35. Let z = x/y, then z > 0. z - 1/z = 24/35. 35z^2 - 24z - 35 = 0. (5z - 7)(7z + 5) = 0. z = 7/5, or z = -5/7 (out as z > 0). Now we have x/y = 7/5, xy = 35. Multiplying together: x^2 = 49. x = 7 or -7. So y = 35/x = 5, or -5. (7,5) and (-7, -5) fit original equations.
Correct and also : The factors of 35 are follows {35,1} or { 7,5) Therefore , which set of numbers satisfied the given instead of providing complicated solutions
@@chuviethuong5081 Yes, it's important to remember (-7,-5). And the context of the test should make it clear whether non-real numbers are allowed. If so, (5i,-7i) and (-5i,7i) are possibilities.
Laughing 😃 😀 😄 you just got your so simple... let y be 7 and x be 5 or vise versa.... lol... love ur formula and remember it can also be 35 and 1 or vise versa
Everyone has own solution for equation. So I understand your solution that showed above TH-cam. But I don't think this solution is the best and easist way to solve the equation and if there is another easier and faster method for this equation, I can't help choosing the method for solving the equation. BECAUSE in case of solving the equation, saving the time to slove equation is the important value.
This can be solve without algebra, but if we eager to use it, x^2y^2=35^2 Let’s say x^2=X, y^2=Y XY=35^2 , X-35^2/X=24 X^2-24X-35^2=0 (X-49)(X+25)=0 X=x^2=49 x=7, 7y=35, y=5 ** x+y=12
and you found 1 solution out of 4. You need to find all 4 values to get perfect score. But I agree this video used way too complicated method to find the solution. set y = 35/x , and insert substitute y of the first equation. Multiplying x^2 on both side, and substitute x^2 with t make it simple quadratic equation.
Из уравнения x2-y2=24получаем x=√24+y2; из уравнения x*y=35 x=35:y, решаем:√24+y2=35:y; получаем y4+24y2-1225=0, решаем квадратное уравнение.Представим что у2=z.тогда получаем уравнение z2+24z-1225=0 отсюда z=49 а значит y=√49=7.Значит x=35:7=5.Без всяких премудростей.Просто и легко.Только не умею число в квадрате писать на телефоне
В ответе есть комплексные числа. Почему-то основная масса комментаторов об этом забывает. Всего 4 ответа: два ответа -- это действительные числа, и два ответа -- это комплексные числа.
Commenters: you made this waaaay too complicated! 😡 Solution: *two real and two imaginary roots, only one of which anyone found by inspection* And that, ladies and gentlemen, is the difference between a Math Olympian and your average youtube commenter
" the difference between a Math Olympian and your average TH-cam commenter ". That's a rather condescending statement I think. Normally in this pedagogic environment it's customary to indicate whether integer, real or complex solutions are required, which if it had been done in this case, should eliminate the " difference " to which you allude.
The equation x*y =35 states that the product of x*y is equal to 35 =7*5 - that is the product of two prime numbers. This is the only pair of natural numbers whose product is 35. No other pair of natural numbers will have the same product. The solution set for x can be expanded to include negatives, complex conjugates and their negatives, -> {x} = { 7,-7,5*I,-5*I } The solution set for y will be {y} = { 5,-5,-7*I, 7*I } The resulting solution set for x+y -> { x+y } = { 12,-12,-2*I,2*I } Thus applying some simple reasoning leads to a shorter solution than applying algebra in its abstract form.
For those interested in a general solution : Given that x*y = p1*p2 where p1,p2 are two different prime numbers, the solution sets are : {x} = { p1, -p1, p2*I, -p2*I } {y} = { p2, -p2, -p1*I, pi*I } { x + y } = { p1+p2, -( pi+p2 ), ( p2-p1)*I, -( p2-p1) *I }
@@joec1920 In the general case, you cannot assume that. But the point here is that a pair of prime numbers works, and using them will get you to the solution much faster. And this is an Olympiad question. The point of the Olympiad is to test cleverness, not to require students to grind through long, plodding algebraic computations.
@@rickdesper I guess it's just my preference, but I prefer a systematic way than a test hack. I care more about understanding the process than speed, but I can see why speed is important in some math competition.
Hello Sir Don’t you think it would be easier to substitute either x or y from (2) and then solve a quadratic equation giving you the solutions … both real (indeed integers) and imaginary? Regards Erick
No need for the quadratic equation factorise it into two factors then find the possible combination remember in these questions the answer is mostly integral
I would look at xy = 35, x^2 - y^2 = 24, and try guessing that x,y are factors of 35 with x > y. x=35 is clearly ridiculous, so I'd move on to x=7,y=5 and it works. So, x + y = 12. Are there other possibilities? Well, if I multiple both x and y by (-1) I get to the same place, so x=-7,y=-5 is also a solution. The thing is: the approach given here is plodding and systematic and will work regardless of what the constants are. But if you don't know this method, it's not something that will quickly occur to a problem solver to use. Whereas a quick guess-and-check takes very little time. And it's just as valid. Now if the constants were different, I could see the problem solver being forced to take the plodding approach. But when the constants lead to an obvious solution, I'd say that the test designers want the test takers to take the obvious path. These exams are timed and finding fast solutions is definitely encouraged. Finally, I think it's a huge waste of time to look at x^2 + y^2 = -74. In tests like these, non-real solutions are not considered unless explicitly asked for.
for x, y are real numbers, xy=35 => xy=5*7=1*35 =(-5)*(-7)=(-1)*(-35) => (x,y)=(5,7)or (7,5) or (1,35) or (35,1) or (-5,-7),(-7,-5) or (-1,-35) or (-35,-1), sub these possible solutions into x*x-y*y=24, we find that the solutions are (7,5) or (-7,-5), therefore x+y=12 or -12.
Just put the valu of Y from second expression and evaluate. You will ge x4 -24x2-35-1225. Now put x2 as a and you will get a simle quadratic equation a2-24a-1225. Thus on solving we get a=49 and -25 or x2= 49 or -25. Thus, u will get two real and two imaginary roots.
I would say first you set X=x+y and Y=x-y. Then you will have XY=24 and X^2 - Y^2=140 and then solve for X^2 by substiuting Y=24/X into the latter. You will have X^2= 144 or -4. This way you can reduce the amount of calculation.
An elegant way to solve this system is to use complex numbers to avoid solving quadratic equation twice. Here we go. Considering x and y as Real numbers. Set z=x+iy so z^2=x^2-y^2+2ixy=24+70i Then |z^2|=√(24^2+70^2)=74 Using z_=x-iy conjugate of z, We have z+iz_=x+iy+ix-i^2y So z+iz_=(x+y)(1+i) Then |z+iz_|^2=2|x+y|^2 Moreover, |z+iz_|^2=|(z+iz_)^2| So 2|x+y|^2=|z^2-z_^2+2izz_| =|z^2-z^2_+2i|z|^2| =|2i.Im(z^2)+2i|z|^2| =|2i.(Im(z^2)+|z|^2| =|2i|.|Im(z^2)+|z|^2| =2.|Im(z^2)+|z|^2| As Im(Z)= imaginary part and. |Z| module of complex Z. Then |x+y|^2=|Im(z^2)+|z^2|| So x+y=+/-√|Im(z^2)+|z^2|| Knowing that z^2=24+70i and |z^2|=74 then x+y=+/-√(70+74)=+/-√144 So x+y=+/-12 Greetings
Can be done this way :- xy = 35 or 4xy = 140 or (x+y)^2 - (x-y)^2 = 140 multiply both side by (x +y)^2 Gives - (x+y)^4 - (x^2-y^2)^2 = 140(x+y)^2 Put (x+y)^2 = t and get t^2 - (24)^2 = 140t or t^2 - 140t - 576 = 0 Solving this quadratic eqn we get t = 144 or -4 Negative value should be avoided so t = 144 or (x+y)^2 = 144 or (x+y) = 12 or -12
How is this problem an Olympiad Algebra Problem? I saw this kind of problems all the time in Pre-cal textbooks I used to teach. From Eq. 2, we see that y = 35/x and sub this in Eq. 1, we have x^2 - (35/x)^2 = 24. Multiply both sides by x^2, we have x^4 - 35^2 = 24x^2 and hence x^4 - 24x^2 - 35^2 = 0. Notice that 35^2 = 5^2 * 7^2 = 25 * 49 and we can factor the quartic equation as (x^2 - 49)(x^2 + 25)= 0. Notice that x^2 + 25 can't be 0, so we have x^2 - 49 = 0, which yields x = 7 or -7. If x = 7, then y = 5, and we have x + y = 12. If x = -7, then y = -5, and we have x + y = -12. The above solution assumes the solutions are real numbers only. If the solutions can be complex numbers, then set x^2 + 25 equal to 0 and obtain x^2 = -25 and hence x = 5i or -5i. If x = 5i, then y = -7i (since xy = (5i)(-7i) = -35i^2 = 35). So x + y = -2i. If x = -5i, then y = 7i and hence x + y = 2i.
A fast way is to use the Pythagorean triple: (x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2 => (x^2 + y^2) = 74 (x + y)^2 = x^2 + y^2 + 2xy = 74 + 2*35 => x + y = ± 12 However, this assumed the equation has a solution at all. So we have to use the Vieta's formula to solve for x,y, knowing their sum and product and |x| > |y|. There are 2 solutions: (7,5) & (-7, -5) so (x+y) = ± 12
There are at least 2 much easier ways to solve the issue. If you are familiar with trigonometry, you may simply substitute x=sqrt(24)*sec(theta) and y=sqrt(24)*tan(theta) and get 24sin(theta)/(1-sin^2(theta))=35 by multiplying the two. Then you can easily solve it to sin(theta)=5/7 or --7/5 (discarded due to abs(sin(theta))=0), then t=12 or -12.
@@varanov7508 ну не скажите. Он искал сумму х+у, не вычисляя каждую переменную отдельно. Я тоже не стала в лоб решать. Сделала преобразования, обозначила х+у переменной "а" и решила квадратное уравнение.
If X=×^2 and X=-y^2, then X+Y=24 and XY=-5^2=-1225. Il follows that X and Y are the solution to T^2-24T-1225=0.The reduced discriminant is 12^2+1225=144+1225=1369=37^2. There are two solutions (×,y) since x and y have the samedi sign because xy is positive: (7,5) and (-7,-5).
It's given that a² - b² = 24 which implies that (a, b) both must be +ve or -ve since the difference of thier squares is a positive integer. The fact that xy = 35 also supports the above statement. Now, since a² - b² = 24 can only be possible if a² is 49 and b² is 25...hence a must be 7 or -7 and b must be 5 or - 5. Idk if there's any crack here but instead of solving it, I tried to observe it and use logic. This method is good for olympiad or tests when time is running out. I have always hated algebraic manipulation because a wrong approach can make the solution lengthy and then if we a stuck somewhere, then we have to develop an approach again only to see that the solution got much more lengthy
Автор - это классика звездеца :) Когда немного поумневший тупарь видит, что стал лучше окружающих - и в итоге мнит себя чуть ли не гением. Ещё вариант - подсобрать лайки/дизлайки/комментарии. Лучшее решение - удалять такие видео из рекомендаций!
Взять пример для устного разбора по программе 5-го класса, довести его решение до полного маразма и после этого хотеть, чтобы дети знали, понимали математику и имели о ней представление? х*у= 35; Раскладываем 35 на множители, вариант всего один: 7*5=35; Далее записываем так: Пусть х=7, у=5, Тогда: 7*5=35; Следовательно: 7^2-5^2=49-25=24; Тогда: Х+У=7+5=12 Любой пример имеет несколько вариантов решения и при выборе решения приветствуется тот, который облегчает, а не усложняет решение. Это для данного примера решение, но если та же система уравнений будет иметь слишком большие числа или иметь усложнённый вид, тогда, в качестве способа решения таких систем, можно использовать предложение высшей математики и заняться вот таким трёхэтажным решением.
This to me is an easier option. I’m not sure if this is a better solution but it is another option: X^2 - Y^2 = 24 Xy = 35 X=35/y (35/y)^2 - y^2 = 24 1225 - y^4 = 24y^2 -y^4 - 24y^2 + 1225 = 0 Let h = y^2 -h^2 - 24h + 1225 = 0 If h = -b +-(b^2 - 4ac)^0.5 / 2a Then h = 24 +-(576 + 4x1225)^0.5/-2 h = 24 +- (576+4900)0.5 / -2 h = 24 +- (74) / -2 h = 98 /-2 h = -49 Y = +- 7i Or h = -50 / -2 h = 25 Y = +- 5 X = 35/y X = 35/5 = 7 (7,5) X+ y = 12 X = 35/-5 = -7 (-7,-5) X+y = -12 X = 35/7i = -5i (-5i,7i) X+y = 2i X = 35/-7i = 5i (5i,-7i) X+y = -2i
1) X와Y가 실수일 때 직관적으로 X=7,Y=5 or X=-7, Y=-5 가 보이지만 굳이 식을 이용해서 풀자면 XY=35 그러므로 Y=35/X X제곱-Y제곱=24에 대입 X제곱-(35/X)의 제곱=24 양변에 X의 제곱을 곱해서 정리하면 X네제곱 - 24×X의제곱 - (35)의 제곱=0 X의 제곱을 T로 치환(T는 0보다 커야함) T의 제곱 - 24T -(35)의 제곱=0 인수분해하면 (T-49)(T+25)=0 T=49 or T=-25 T는 0보다 커야함으로 T=49 그러므로 X의 제곱은 49 X=7 or -7 이것을 X의 제곱-Y의제곱=24에 대입 X=7이면 Y=5(X와Y의 곱이 양수이므로 서로 부호가 같아야함으로) X=-7이면 Y=-5 두 식에서 XY=35를 만족하는 것은 X=7 ,Y=5 or X=-7, Y=-5 그러므로 X+Y=12 or X+Y=-12 2) X와Y가 둘 다 허수일 때 X를 Ai로, Y를 Bi로 치환해서 다시 연립해서 계산해보면 X=5i 일 때 Y= -7i 따라서 X+Y= - 2i X= - 5i 일 때 Y= 7i 따라서 X+Y= 2i 그러므로 X+Y 값으로 가능한 것은 12, -12 , 2i , -2i
If X and Y are both real numbers, XY=35 therefore Y=35/X X squared - Y squared = 24 X squared - (35/X) squared = 24 Multiplying both sides by X squared : X to the power of 4 - 24×X squared - 35 squared=0 Replacing X squared with T ( T must be greater than 0) T squared - 24T - 35 squared=0 If factoring (T-49)(T+25)=0 T=49 or T=-25 So T=49 Therefore, X squared is 49. X=7 or -7 (X squared - Y squared = 24) If X=7, then Y=5 (because XY is greater than 0 , so they must have the same sign) If X=-7 then Y=-5 Which of the two equations satisfies XY=35 Therefore X=7 ,Y=5 or X=-7, Y=-5 X+Y=12 or X+Y=-12 If X and Y are both imaginary numbers, X->Ai, Y->Bi('A' is real number, 'i' is imaginary number) 1) (Ai) squared - (Bi) squared =24 Therefore - A squard + B squard=24 2) Ai × Bi =35 Therefore - AB =35 ( so, B= - 35/A) -A squared + (- 35/A) squared=24 By the same way A to the power of 4 + 24×A squared -35 squared =0 Replacing A squared with T T squared + 24T - 35 squared=0 (T-25)(T+49)=0(T must be greater than 0) Therefore T=25 A=5 or -5 If A=5 then B=-7 If A=-5 then B =7 Therefore if X =5i, Y= -7i then X+Y= -2i If X=-5i, Y=7i then X+ Y= 2i In conclusion, X+Y=12, -12, 2i, -2i
I watched your video, it's very good but I have 1 question: According to the data (2), x.y = 35 . Therefore (x.x) + (y.y) > 0 . So (x.x) + (y.y) = 74 . Impossible = -74. Furthermore, the given data is missing: x,y belongs to n or R . If x,y belong to n, then we find x = 7 and y = 5. So it's simply: x + y = 12 . But it doesn't have to be that complicated.
"Solving" a question like this means finding ALL the solutions and proving that there are no others. Just looking at it and saying "umm, 7 and 5 work" is not solving it.
Let me try this in my way...(x^2-y^2)^2=(x^2+y^2)^2-4(xy)^2...hence, (x^2+y^2)=√(24^2+4×35^2)=74. Now (x+y)^2=(x^2+y^2)+2xy=74+70=144. x+y=+-12, here I confess a thing, I knew about √-1=i, but couldn't realise that the 74 is actually a squareroot, and it would lead to two more complex solutions of this problem... otherwise, it isn't too hard at all.
there's only 2 pairs of whole numbers that when multipled by each other, gives 35: 1 x 35 7 x 5 1 x 35 doesn't satisfy the first equation, so.................
Да тут и решать ничего не надо. Мы имеем целые числа. Если xy=35, то только может быть один вариант:7×5=35. Х=7, у=5. Х не может быть равен 5, а у не может быть равен 7,потому что тогда не выполняется равенство x в квадрате плюс у в квадрате равно 24. Вот и все. 7 плюс 5 равно 12.
I love how people say 7 and 5 and think that they are olympiad level while giving the wrong answer, their self love and bloating is on another level i mean reaching this level is hard af good job
Why didn't you just solve xy=35 for say y to get y=35/x then substitute that into the top equation x^2-y^2=24 to get x^4-24x^2-1225=0. Solve this for x. You will get 4 values of x, which you can then sub into either equation to get 4 values of y. You'll arrive at the same solution. This method is significantly easier. Most students will know how to factor a 4th degree polynomial
you can use substitute method by converting into quadratic equation. it will take less time than your method. there is not enough time in exam center to solve by your method.
@@madurappankalyanaraman8015 because X=-7, y=-5 also works in the real number domain, and then the presented solution assumes the complex number domain, which adds two more solutions
just find for the xy first bc we know xy = 35 and 35 only has the factor of 7 and 5 then we can input 7(5) = 35, since x > y then 7^2 - 5^2 = 24 then x + y = 7 + 5 = 12
@@igorfujs7349 the x is -7 and y = -5, the it would be 2 the result. can be use in the solution but in the addition part it would be a total different result
Look it is very simple I solve it without algebraically look in que xy=35 We know that factor of 35 are 1,5,7,35 If by trial and error method let x=35and y=1 Then (35)²-(1)²=1224 so We take another value X=7 and y=5 Then we get (7)²-(5)²=24 Which is satisfied to the 1st EQ of question Then x+y=7+5=12
@learncommunolizer! Oh the name suggests a big larger than life personality is about to appear! x2-y2=24 => x2>y2 => x>y xy=35 xy=5*7 both are prime Obviously 1*35 is not ok. x=7; y=5 => x+y = 7+5 = 12
Please note that sqrt 4 is not +/- 2. sqrt 4 is the positive square root of 4, which is equal to 2 only. So when x^2 = 4, you should write x = +/- sqrt 4 = +/- 2, instead of writing x = sqrt 4 = +/- 2. Think of when you are asked to evaluate sqrt 4 + sqrt 9. The answer is 2 + 3 = 5. I don't think the answer is (+/-2) + (+/-3).
OMG I seriously would never use this way to solve. I would just add up after *2(2) to make (x-y)^2 then solve for x+y later on. You basically put my soul in eternal torment.
I think we can solve it in a simpler way. you can substitute the values x = 7 and y = 5 or x = -7 dan y = - 5. The results are in accordance with the equation requested. so the final result is 12 or -12. Am I right?
Let z=x+iy, then z^2=(x^2-y^2)+2i(xy)=24+70i so z =+-z0 where z0=7+5i is a racine of 24+70i which can be found by taking the polar form of 24+70i. then x+y=+-12 and that's all!!!
tell me is it correct or not, Here xy = 35 So, X = 35/y Now puting the value of y in x^2 - y^2 or (x+y) (x-y) = 24 Now, (35/y + y) ( 35/y -y) = 24 Or, [(35+y^2)/y] [(35-y^2)/y] = 24 Now getting LCM [(35+y^2)(35-y^2)]/y^2 = 24 Or, (35^2 - y^2)/y^2 = 24 Or, 1225 - y^2 = 24y^2 Or, 1225 = 24y^2 + y^2 Or, 1225 = 25y^2 Or, 1225/25 = y^2 Or, 49 = y^2 So, y = 7 puting the value of y=7 in xy = 35 Now, x7 = 35 So, x = 35/7 = 5 Now, (x+y) = (5+7) = 12 Can't we do it
Well not a person really good at maths but If xy=35, then either x is 7or 5 or -7 or -5 cause 35 is only divisible by these two , so value of x+y is -12 or 12
This is the most complicated way to solve this problem. The simple way is to use equation 2 to to get y=(35/x) and substitute it into equation 1. With a little restructuring, you get a quadratic polynomial and use the quadratic equation to solve for x. Once you have the values of x, it is easy to get the values of y.
In fact you have a system of 2 equations in 2 unknowns so it's complete
Just do it by inspection; no Quadratic Equation or even square roots to mess with.
1) Factor out x² - y² = 24.
(x + y)(x - y) = 24
2) First, find the solution for only positive real integers.
List the possible factors for 24 (in order). Note that since 24 > 0, x > y.
(x + y)
6 8 12 24
(x - y)
4 3 2 1
i.e., [(x +y),(x-y)] = (6,4), (8,3), (12,2), and (24,1).
3) List the values of 2x & 2y for these ordered pairs.
(x + y) + (x - y) = 2x
(x + y) - (x - y) = 2y
2x
10 11 14 25
2y
2 5 10 23
4) List the values of 4xy = (2x)(2y).
4xy
20 55 140 575
5) The third value (140) yields the given initial condition.
4xy = 140
xy = 35
6) The third pair of factors is the solution for positive real integers.
(x + y) = 12
(x - y) = 2
x = 7
y = 5
7) Negative (real) integers provide another branch.
(x + y) = - 12
(x - y) = - 2
x = - 7
y = - 5
or
(x,y) = (7,5) or (- 7,- 5)
and (x + y) = ± 12
8) For complex solutions, note that 24 and 35 are explicitly real, so complex values for x and y can only be explicitly imaginary, i.e.,
x = i u
y = i v
9) Factor out the original expression in terms of u and v.
x² - y² = - u² + v² = 24
or
(- u + v)(u + v) = 24
10) Use the method above used for real numbers to arrive at the solution for u and v.
(|u|,|v|) = (5,7)
11) To find the quadrant of the imaginary solution, recall the condition
xy = 35
The only way the product of two imaginary numbers can yield a positive value (i.e., 35) is if their signs are complementary. In other words,
(5i)(- 7i) = + 35
and
(- 5i)(7i) = +35
Thus, the imaginary solution is
(x,y) = (5i,- 7i) or (- 5i,7i)
and (x + y) = ± 2i
@@ScrewFlanders You do realize that this approach is infinitely more complicated than substitution followed by the quadratic formula, right?
@zachowland Well, you're entitled to your own opinion. I solved the problem in a couple minutes using this method (without watching the video first), so I'm not sure I accept that my method is "infinitely" more complicated. In _my_ opinion, my method is simpler. 😉
@@ScrewFlanders Look at how long it took you to explain it in comparison. It is not a matter of opinion.
From x^2 - y^2 = 24. Divide it by xy = 35. (x/y) - (y/x) = 24/35. Let z = x/y, then z > 0. z - 1/z = 24/35. 35z^2 - 24z - 35 = 0.
(5z - 7)(7z + 5) = 0. z = 7/5, or z = -5/7 (out as z > 0).
Now we have x/y = 7/5, xy = 35. Multiplying together: x^2 = 49. x = 7 or -7. So y = 35/x = 5, or -5. (7,5) and (-7, -5) fit original equations.
XY=35, Itself solving the problem. X,Y {7,5}. Other equation can only help to bind the value with respective variables only.
Автор демонстрирует идиотское решение! Графомания!
В условие задачи, решение очевидно.
XY=35=7×5, X+Y=12, 49-25=24.
You missed {-7,-5} result, too. And in the equation 7x5 = 5x7 cannot applying to x = 5, -5 and y = 7, -7.
Correct and also :
The factors of 35 are follows
{35,1} or { 7,5)
Therefore , which set of numbers satisfied the given instead of providing complicated solutions
@@chuviethuong5081 Yes, it's important to remember (-7,-5). And the context of the test should make it clear whether non-real numbers are allowed. If so, (5i,-7i) and (-5i,7i) are possibilities.
Laughing 😃 😀 😄 you just got your so simple... let y be 7 and x be 5 or vise versa.... lol... love ur formula and remember it can also be 35 and 1 or vise versa
You turned a simple solution into complicated one
Is it that hard to acknowledge that everyone has a different way of solving things? Grow up
I gotta agree with u
Everyone has own solution for equation. So I understand your solution that showed above TH-cam. But I don't think this solution is the best and easist way to solve the equation and if there is another easier and faster method for this equation, I can't help choosing the method for solving the equation. BECAUSE in case of solving the equation, saving the time to slove equation is the important value.
I agree with you !
But a *complete* answer.
Sir, you have made a simple problem, into very complicated one.
But this method will work no matter what numbers are on the RHS
This can be solve without algebra, but if we eager to use it,
x^2y^2=35^2
Let’s say x^2=X, y^2=Y
XY=35^2 , X-35^2/X=24
X^2-24X-35^2=0
(X-49)(X+25)=0
X=x^2=49
x=7, 7y=35, y=5
** x+y=12
Don't forget simply inserting a few integers to see if anything fits - took only a few seconds to find x=7 and y=5!
Then you get zero point in this case ~ no solving process
This is about "How to solve".... not about "solving"😂😂😂
and you found 1 solution out of 4. You need to find all 4 values to get perfect score. But I agree this video used way too complicated method to find the solution. set y = 35/x , and insert substitute y of the first equation. Multiplying x^2 on both side, and substitute x^2 with t make it simple quadratic equation.
Из уравнения x2-y2=24получаем x=√24+y2; из уравнения x*y=35 x=35:y, решаем:√24+y2=35:y; получаем y4+24y2-1225=0, решаем квадратное уравнение.Представим что у2=z.тогда получаем уравнение z2+24z-1225=0 отсюда z=49 а значит y=√49=7.Значит x=35:7=5.Без всяких премудростей.Просто и легко.Только не умею число в квадрате писать на телефоне
В ответе есть комплексные числа. Почему-то основная масса комментаторов об этом забывает.
Всего 4 ответа: два ответа -- это действительные числа, и два ответа -- это комплексные числа.
@@hydraoni0n так я ж написала корень квадратный √
@@hydraoni0n да, вы правы
Интерес вызывает только комплексные числа в решении. Плюс/минус 7 и 5 - очевидно вычисляются в уме
^ это знак степени. Он у всех есть на телефоне.
Мне 68 лет, и я - не математик,но за минуту решила разными способами.Я молодец
Commenters: you made this waaaay too complicated! 😡
Solution: *two real and two imaginary roots, only one of which anyone found by inspection*
And that, ladies and gentlemen, is the difference between a Math Olympian and your average youtube commenter
*PERFECTLY SAID!*
" the difference between a Math Olympian and your average TH-cam commenter ". That's a rather condescending statement I think. Normally in this pedagogic environment it's customary to indicate whether integer, real or complex solutions are required, which if it had been done in this case, should eliminate the " difference " to which you allude.
@@crustyoldfart When not specified, assume the largest set, that is Complex. That is the rule in maths.
The equation x*y =35 states that the product of x*y is equal to 35 =7*5 - that is the product of two prime numbers.
This is the only pair of natural numbers whose product is 35. No other pair of natural numbers will have the same product.
The solution set for x can be expanded to include negatives, complex conjugates and their negatives,
-> {x} = { 7,-7,5*I,-5*I }
The solution set for y will be {y} = { 5,-5,-7*I, 7*I }
The resulting solution set for x+y -> { x+y } = { 12,-12,-2*I,2*I }
Thus applying some simple reasoning leads to a shorter solution than applying algebra in its abstract form.
For those interested in a general solution :
Given that x*y = p1*p2 where p1,p2 are two different prime numbers, the solution sets are :
{x} = { p1, -p1, p2*I, -p2*I }
{y} = { p2, -p2, -p1*I, pi*I }
{ x + y } = { p1+p2, -( pi+p2 ), ( p2-p1)*I, -( p2-p1) *I }
you just assumed that x and y must be natural numbers? It's true in this case, but I can't see how you can assume that.
@@joec1920 In the general case, you cannot assume that. But the point here is that a pair of prime numbers works, and using them will get you to the solution much faster. And this is an Olympiad question. The point of the Olympiad is to test cleverness, not to require students to grind through long, plodding algebraic computations.
@@rickdesper I guess it's just my preference, but I prefer a systematic way than a test hack. I care more about understanding the process than speed, but I can see why speed is important in some math competition.
Ummmm, 1x35 is still 35 and both are natural numbers too. Not that it isn't obvious which ones are right, but there are two other factors to consider.
Hello Sir
Don’t you think it would be easier to substitute either x or y from (2) and then solve a quadratic equation giving you the solutions … both real (indeed integers) and imaginary?
Regards
Erick
Both are imaginery nos
My thinking too
I actually got it by your method. I came to a quadratic equation.
我的做法更簡單。
No need for the quadratic equation factorise it into two factors then find the possible combination remember in these questions the answer is mostly integral
Your technique solving this problem that makes me refresh my math lesson I learned 62 years ago! Thanks!
Excellent! Glad to help!👍
Do a replacement of x^2 with U. U equals 49 or -25. Then you can get the answers of x. A lot simple.
I would look at xy = 35, x^2 - y^2 = 24, and try guessing that x,y are factors of 35 with x > y. x=35 is clearly ridiculous, so I'd move on to x=7,y=5 and it works. So, x + y = 12. Are there other possibilities? Well, if I multiple both x and y by (-1) I get to the same place, so x=-7,y=-5 is also a solution.
The thing is: the approach given here is plodding and systematic and will work regardless of what the constants are. But if you don't know this method, it's not something that will quickly occur to a problem solver to use. Whereas a quick guess-and-check takes very little time. And it's just as valid.
Now if the constants were different, I could see the problem solver being forced to take the plodding approach. But when the constants lead to an obvious solution, I'd say that the test designers want the test takers to take the obvious path. These exams are timed and finding fast solutions is definitely encouraged.
Finally, I think it's a huge waste of time to look at x^2 + y^2 = -74. In tests like these, non-real solutions are not considered unless explicitly asked for.
Любила в школе решать. Помню до сих пор. Школу окончила 53 года назад.
for x, y are real numbers, xy=35 => xy=5*7=1*35 =(-5)*(-7)=(-1)*(-35) => (x,y)=(5,7)or (7,5) or (1,35) or (35,1) or (-5,-7),(-7,-5) or (-1,-35) or (-35,-1), sub these possible solutions into x*x-y*y=24, we find that the solutions are (7,5) or (-7,-5), therefore x+y=12 or -12.
Just put the valu of Y from second expression and evaluate.
You will ge
x4 -24x2-35-1225.
Now put x2 as a and you will get a simle quadratic equation
a2-24a-1225.
Thus on solving we get a=49 and -25 or x2= 49 or -25.
Thus, u will get two real and two imaginary roots.
I would say first you set X=x+y and Y=x-y. Then you will have XY=24 and X^2 - Y^2=140 and then solve for X^2 by substiuting Y=24/X into the latter. You will have X^2= 144 or -4. This way you can reduce the amount of calculation.
An elegant way to solve this system is to use complex numbers to avoid solving quadratic equation twice. Here we go. Considering x and y as Real numbers. Set z=x+iy so z^2=x^2-y^2+2ixy=24+70i
Then |z^2|=√(24^2+70^2)=74
Using z_=x-iy conjugate of z,
We have z+iz_=x+iy+ix-i^2y
So z+iz_=(x+y)(1+i)
Then |z+iz_|^2=2|x+y|^2
Moreover, |z+iz_|^2=|(z+iz_)^2|
So 2|x+y|^2=|z^2-z_^2+2izz_|
=|z^2-z^2_+2i|z|^2|
=|2i.Im(z^2)+2i|z|^2|
=|2i.(Im(z^2)+|z|^2|
=|2i|.|Im(z^2)+|z|^2|
=2.|Im(z^2)+|z|^2|
As Im(Z)= imaginary part and. |Z| module of complex Z.
Then
|x+y|^2=|Im(z^2)+|z^2||
So
x+y=+/-√|Im(z^2)+|z^2||
Knowing that z^2=24+70i and |z^2|=74 then
x+y=+/-√(70+74)=+/-√144
So
x+y=+/-12
Greetings
When x^2+y^2, there should have no negative numbers, the answer of -74 should be rejected.
This is where the two imaginary solutions come from..
Well...the instructions at some point should make it clear whether non-real complex numbers are being considered. By default, they usually are not.
Can be done this way :-
xy = 35
or 4xy = 140
or (x+y)^2 - (x-y)^2 = 140
multiply both side by (x +y)^2
Gives -
(x+y)^4 - (x^2-y^2)^2 = 140(x+y)^2
Put (x+y)^2 = t and get
t^2 - (24)^2 = 140t
or t^2 - 140t - 576 = 0
Solving this quadratic eqn we get t = 144 or -4
Negative value should be avoided so
t = 144
or (x+y)^2 = 144
or (x+y) = 12 or -12
x=7, y=5.
x+y=12
How is this problem an Olympiad Algebra Problem? I saw this kind of problems all the time in Pre-cal textbooks I used to teach. From Eq. 2, we see that y = 35/x and sub this in Eq. 1, we have x^2 - (35/x)^2 = 24. Multiply both sides by x^2, we have x^4 - 35^2 = 24x^2 and hence x^4 - 24x^2 - 35^2 = 0. Notice that 35^2 = 5^2 * 7^2 = 25 * 49 and we can factor the quartic equation as (x^2 - 49)(x^2 + 25)= 0. Notice that x^2 + 25 can't be 0, so we have x^2 - 49 = 0, which yields x = 7 or -7. If x = 7, then y = 5, and we have x + y = 12. If x = -7, then y = -5, and we have x + y = -12. The above solution assumes the solutions are real numbers only. If the solutions can be complex numbers, then set x^2 + 25 equal to 0 and obtain x^2 = -25 and hence x = 5i or -5i. If x = 5i, then y = -7i (since xy = (5i)(-7i) = -35i^2 = 35). So x + y = -2i. If x = -5i, then y = 7i and hence x + y = 2i.
A fast way is to use the Pythagorean triple:
(x^2 + y^2)^2 = (x^2 - y^2)^2 + (2xy)^2
=> (x^2 + y^2) = 74
(x + y)^2 = x^2 + y^2 + 2xy = 74 + 2*35
=> x + y = ± 12
However, this assumed the equation has a solution at all. So we have to use the Vieta's formula to solve for x,y, knowing their sum and product and |x| > |y|. There are 2 solutions: (7,5) & (-7, -5) so (x+y) = ± 12
I haven't seen such a difficult way to solve a simple question
He is asking for a solution, not estimation.
But his solution works for any choice of number on the right
There are at least 2 much easier ways to solve the issue. If you are familiar with trigonometry, you may simply substitute x=sqrt(24)*sec(theta) and y=sqrt(24)*tan(theta) and get 24sin(theta)/(1-sin^2(theta))=35 by multiplying the two. Then you can easily solve it to sin(theta)=5/7 or --7/5 (discarded due to abs(sin(theta))=0), then t=12 or -12.
Hay. Xem và nhớ lại thời học sinh, vẫn nhớ và vận dụng các hằng đẳng thức đáng nhớ. Cảm ơn bạn ❤️
Thanks and Your welcome!🙏❤️
Выражаешь Х через У или наоборот и решаешь обычное простейшее квадратное уравнение ... Тут нагорожен целый огород - непонятно зачем ...
Очевидно, он не хотел в лоб решать.
@@AlexandraMarchenkova очевидно он хотел повыпендриваться))
@@varanov7508 ну не скажите. Он искал сумму х+у, не вычисляя каждую переменную отдельно. Я тоже не стала в лоб решать. Сделала преобразования, обозначила х+у переменной "а" и решила квадратное уравнение.
@@AlexandraMarchenkova ну так можно Х выразить через У и решить потом тоже квадратное уравнение нафиг в дебри лезть? Тут уровень 9 класса)
@@varanov7508 часто задачу можно решить несколькими способами. Можно в лоб, а можно сделать это более интересным способом. Что он и сделал.
There is a mistake : square root of something is a positive value. There are no two cases with +74 and -74. Only +74
If X=×^2 and X=-y^2, then X+Y=24 and XY=-5^2=-1225. Il follows that X and Y are the solution to T^2-24T-1225=0.The reduced discriminant is 12^2+1225=144+1225=1369=37^2. There are two solutions (×,y) since x and y have the samedi sign because xy is positive: (7,5) and (-7,-5).
-5^2=-1225?*
Xy=35
35=7×5or5×7
(X×X)-(Y×Y)=24
Ex:X=7,Y=5
(7×7)-(5×5)=24
49-25=24
X+Y=?
7+5=12
It's given that a² - b² = 24 which implies that (a, b) both must be +ve or -ve since the difference of thier squares is a positive integer. The fact that xy = 35 also supports the above statement.
Now, since a² - b² = 24 can only be possible if a² is 49 and b² is 25...hence a must be 7 or -7 and b must be 5 or - 5.
Idk if there's any crack here but instead of solving it, I tried to observe it and use logic. This method is good for olympiad or tests when time is running out. I have always hated algebraic manipulation because a wrong approach can make the solution lengthy and then if we a stuck somewhere, then we have to develop an approach again only to see that the solution got much more lengthy
Тут в лоб решается быстрее чем вы расписываете. Единственная сложность посчитать sqrt (5476). Даже с учётом биквадратного уравнения.
Я решил тоже по - другому и проще : через биквадратное уравнение.
Автор - это классика звездеца :)
Когда немного поумневший тупарь видит, что стал лучше окружающих - и в итоге мнит себя чуть ли не гением.
Ещё вариант - подсобрать лайки/дизлайки/комментарии.
Лучшее решение - удалять такие видео из рекомендаций!
Взять пример для устного разбора по программе 5-го класса, довести его решение до полного маразма и после этого хотеть, чтобы дети знали, понимали математику и имели о ней представление?
х*у= 35;
Раскладываем 35 на множители, вариант всего один: 7*5=35;
Далее записываем так:
Пусть х=7, у=5,
Тогда: 7*5=35;
Следовательно:
7^2-5^2=49-25=24;
Тогда:
Х+У=7+5=12
Любой пример имеет несколько вариантов решения и при выборе решения приветствуется тот, который облегчает, а не усложняет решение.
Это для данного примера решение, но если та же система уравнений будет иметь слишком большие числа или иметь усложнённый вид, тогда, в качестве способа решения таких систем, можно использовать предложение высшей математики и заняться вот таким трёхэтажным решением.
This to me is an easier option.
I’m not sure if this is a better solution but it is another option:
X^2 - Y^2 = 24
Xy = 35
X=35/y
(35/y)^2 - y^2 = 24
1225 - y^4 = 24y^2
-y^4 - 24y^2 + 1225 = 0
Let h = y^2
-h^2 - 24h + 1225 = 0
If h = -b +-(b^2 - 4ac)^0.5 / 2a
Then
h = 24 +-(576 + 4x1225)^0.5/-2
h = 24 +- (576+4900)0.5 / -2
h = 24 +- (74) / -2
h = 98 /-2
h = -49
Y = +- 7i
Or
h = -50 / -2
h = 25
Y = +- 5
X = 35/y
X = 35/5 = 7
(7,5)
X+ y = 12
X = 35/-5 = -7
(-7,-5)
X+y = -12
X = 35/7i = -5i
(-5i,7i)
X+y = 2i
X = 35/-7i = 5i
(5i,-7i)
X+y = -2i
7x5= 35
7^2-5^2=24
7+5=12
Good video, it is a fine solution!
Yes, thanks
1) X와Y가 실수일 때
직관적으로 X=7,Y=5 or X=-7,
Y=-5 가 보이지만
굳이 식을 이용해서 풀자면
XY=35 그러므로 Y=35/X
X제곱-Y제곱=24에 대입
X제곱-(35/X)의 제곱=24
양변에 X의 제곱을 곱해서 정리하면
X네제곱 - 24×X의제곱 - (35)의 제곱=0
X의 제곱을 T로 치환(T는 0보다 커야함)
T의 제곱 - 24T -(35)의 제곱=0
인수분해하면
(T-49)(T+25)=0
T=49 or T=-25
T는 0보다 커야함으로
T=49
그러므로 X의 제곱은 49
X=7 or -7
이것을 X의 제곱-Y의제곱=24에 대입
X=7이면 Y=5(X와Y의 곱이 양수이므로 서로 부호가 같아야함으로)
X=-7이면 Y=-5
두 식에서 XY=35를 만족하는 것은
X=7 ,Y=5 or X=-7, Y=-5
그러므로 X+Y=12 or X+Y=-12
2) X와Y가 둘 다 허수일 때
X를 Ai로, Y를 Bi로 치환해서 다시 연립해서 계산해보면
X=5i 일 때 Y= -7i 따라서 X+Y= - 2i
X= - 5i 일 때 Y= 7i 따라서 X+Y= 2i
그러므로 X+Y 값으로 가능한 것은
12, -12 , 2i , -2i
간단한 방정식을 상당히 복잡하게 해결한다.
Beauty about maths is if follow proper logic, you ends up to correct answer. ❤
Wouldn't substitution be easier? Just asking
You're right. Moreover, substitution is general technique and should be preferred from educational reasons (math is not magic).
If X and Y are both real numbers,
XY=35 therefore Y=35/X
X squared - Y squared = 24
X squared - (35/X) squared = 24
Multiplying both sides by X squared
: X to the power of 4 - 24×X squared - 35 squared=0
Replacing X squared with T ( T must be greater than 0)
T squared - 24T - 35 squared=0
If factoring
(T-49)(T+25)=0
T=49 or T=-25
So T=49
Therefore, X squared is 49.
X=7 or -7
(X squared - Y squared = 24)
If X=7, then Y=5 (because XY is greater than 0 , so they must have the same sign)
If X=-7 then Y=-5
Which of the two equations satisfies XY=35
Therefore X=7 ,Y=5 or X=-7, Y=-5
X+Y=12 or X+Y=-12
If X and Y are both imaginary numbers,
X->Ai, Y->Bi('A' is real number, 'i' is imaginary number)
1) (Ai) squared - (Bi) squared =24
Therefore - A squard + B squard=24
2) Ai × Bi =35
Therefore - AB =35 ( so, B= - 35/A)
-A squared + (- 35/A) squared=24
By the same way
A to the power of 4 + 24×A squared -35 squared =0
Replacing A squared with T
T squared + 24T - 35 squared=0
(T-25)(T+49)=0(T must be greater than 0)
Therefore T=25
A=5 or -5
If A=5 then B=-7
If A=-5 then B =7
Therefore if X =5i, Y= -7i then
X+Y= -2i
If X=-5i, Y=7i then X+ Y= 2i
In conclusion, X+Y=12, -12, 2i, -2i
This is the most inefficient way to solve this equation I have seen so far. KEEP TRYING.
I watched your video, it's very good but I have 1 question: According to the data (2), x.y = 35 . Therefore (x.x) + (y.y) > 0 . So (x.x) + (y.y) = 74 . Impossible = -74. Furthermore, the given data is missing: x,y belongs to n or R . If x,y belong to n, then we find x = 7 and y = 5. So it's simply: x + y = 12 . But it doesn't have to be that complicated.
X=7, Y=5
7+5=12
By inspection, I saw this answer in two seconds. This is too easy for math olympics.
Same
This is not the only solution.
Get real. Was there a 3 dimensional premise?
"Solving" a question like this means finding ALL the solutions and proving that there are no others. Just looking at it and saying "umm, 7 and 5 work" is not solving it.
Let me try this in my way...(x^2-y^2)^2=(x^2+y^2)^2-4(xy)^2...hence, (x^2+y^2)=√(24^2+4×35^2)=74. Now (x+y)^2=(x^2+y^2)+2xy=74+70=144. x+y=+-12, here I confess a thing, I knew about √-1=i, but couldn't realise that the 74 is actually a squareroot, and it would lead to two more complex solutions of this problem... otherwise, it isn't too hard at all.
there's only 2 pairs of whole numbers that when multipled by each other, gives 35:
1 x 35
7 x 5
1 x 35 doesn't satisfy the first equation, so.................
You forgot -7 and -5?
There are several simple methods to do this problem It is a long and complicated one
Да тут и решать ничего не надо. Мы имеем целые числа. Если xy=35, то только может быть один вариант:7×5=35.
Х=7, у=5. Х не может быть равен 5, а у не может быть равен 7,потому что тогда не выполняется равенство x в квадрате плюс у в квадрате равно 24.
Вот и все. 7 плюс 5 равно 12.
I love how people say 7 and 5 and think that they are olympiad level while giving the wrong answer, their self love and bloating is on another level i mean reaching this level is hard af good job
No entiendo porque se demora 10 min para resolverlo. Tan sólo con verlo en dos segundos se obtiene que el resultado es + o - 12. X=+-7 y Y=+-5
x=7; y=5 (by inspection)
What about x = -7 and y = -5? This is quadratic remember! Never one solution.. go back to school and learn some real math
@@legacies9041 why be rude
Why didn't you just solve xy=35 for say y to get y=35/x then substitute that into the top equation x^2-y^2=24 to get x^4-24x^2-1225=0. Solve this for x. You will get 4 values of x, which you can then sub into either equation to get 4 values of y. You'll arrive at the same solution.
This method is significantly easier. Most students will know how to factor a 4th degree polynomial
Interesante ver como resolver usando las bases algebraicas con tantas vueltas...
si me agrada las demostraciones
X=7;y=5
XY=35
x= 35/y
Subst.
(35/y)² - y²=24
1225/y² - y²=24
Multiply the whole equation by y²
1225 - y⁴=24y²
y⁴+24y²-1225
1225= 49x25
= 49 -25=24
(Y²+49)(y²-25)
Y²= -49 y²=25, y=5 solution
Not a sol.
x=35/y
x=35/5
x=7
X+Y=12; X = 7; Y = 5
or X+Y = -12 (X=-7; Y=-5)
you can use substitute method by converting into quadratic equation. it will take less time than your method.
there is not enough time in exam center to solve by your method.
X =7 Y=5
Wrong
@@legacies9041
How you are saying that my answer is wrong?
@@madurappankalyanaraman8015 because X=-7, y=-5 also works in the real number domain, and then the presented solution assumes the complex number domain, which adds two more solutions
Put y=mx,divide both,solve for m…and we get values.short and simple
just find for the xy first bc we know xy = 35 and 35 only has the factor of 7 and 5 then we can input 7(5) = 35, since x > y then 7^2 - 5^2 = 24
then x + y = 7 + 5 = 12
X=-5, y=-7???
@@igorfujs7349 the x is -7 and y = -5, the it would be 2 the result. can be use in the solution but in the addition part it would be a total different result
@@myherosuper you dont math my friend but I dont blame you, it is the education system.
Я решил эту задачку за минуту , без всяких уравнений . Это же очевидно !
как?
Метод подстановки
Выложите своё решение, пожалуйста.
35 = 5х7! Дальше просто логика!😅😅😅
@@kulyanjanabayeva4062 wrong
Look it is very simple I solve it without algebraically look in que xy=35
We know that factor of 35 are 1,5,7,35
If by trial and error method let x=35and y=1
Then (35)²-(1)²=1224 so
We take another value
X=7 and y=5
Then we get
(7)²-(5)²=24
Which is satisfied to the 1st EQ of question
Then x+y=7+5=12
When you need to reach the 10 minute mark 😭😭
@learncommunolizer! Oh the name suggests a big larger than life personality is about to appear!
x2-y2=24
=> x2>y2
=> x>y
xy=35
xy=5*7 both are prime
Obviously 1*35 is not ok.
x=7; y=5
=> x+y = 7+5 = 12
Método del tanteo:
xy = 35, entonces x=7, y=5,
Comprobando: 7x7 - 5x5 = 49 - 25 = 24
Respuesta: x + y = 7 + 5 = 12
No me imagino ponerme a tantear y luego descubrir que el resultado era decimal, no es el caso pero sería gracioso xd
@@Triple_D8 Cuando el tanteo falla hay que usar otras estrategias.
Please note that sqrt 4 is not +/- 2.
sqrt 4 is the positive square root of 4, which is equal to 2 only.
So when x^2 = 4, you should write x = +/- sqrt 4 = +/- 2, instead of writing x = sqrt 4 = +/- 2.
Think of when you are asked to evaluate sqrt 4 + sqrt 9. The answer is 2 + 3 = 5. I don't think the answer is (+/-2) + (+/-3).
OMG I seriously would never use this way to solve. I would just add up after *2(2) to make (x-y)^2 then solve for x+y later on. You basically put my soul in eternal torment.
12
X=35/y
(35/y)²-(y²)²=24
1225-y⁴=24y²
Let y² =k
K²+24k-1225=0
K²+49k-25k-1225=0
K(k+49)-25(k+49)=0
K-25=0 or k+49=0
K=25 or k=-49
~y²=k but k= 25
Y²=25
Y=√25
Y=5
Therefore x=35/y
=35/5
X=7
Therefore
X+y =7+5
=12✓
Calculated by kingflyover mgeni
X²is positive and y² is positive too, then (x²+y²) is a positive number, why should consider the negative when taking the square root?
Oh my! There is a very easy solution, but you made it so complicated. X= 35/Y. Substitute X in the first equation by 35/Y, and that will do the trick.
I think we can solve it in a simpler way. you can substitute the values x = 7 and y = 5 or x = -7 dan y = - 5. The results are in accordance with the equation requested. so the final result is 12 or -12. Am I right?
12. One minute by trying in my head. 7sq-5sq=24; 7*5=35; 7+5=12
It is the third time in a row that i watch one of your videos, i love it!
Glad you enjoy it!
Смешно.
xy =35, 7x5=35, 7sq =49 - 5sq=25 49-25=24. x+y = 7+5'12
(x^2 + y^2)^2
= (x^2 - y^2)^2 + 4•[(xy)^2]
= 576 + 4900
= 5476
= 74^2
= [(x + y)^2 - 70]^2
so we have
0 = [(x + y)^2 - 70 - 74]•[(x +y)^2 - 70 + 74]
x + y = 12 or - 12 and 2i or -2i
.x=y/35
.y^2/1225-y^2=24.
.y^2(1-1225)/1225=24
.y^2=(24×1225)/(-1224)
(.x+y)^2=x^2+70-y^2+2y^2=
=94+48×1225/1224=142
.x+y=~12
Simple ko aur hard kar Diya. Bachhe ko kaise पता chalega ki kab apni taraf se kuch addition ya deletion karna होगा ?
His method here may seem to be longer, but it works no matter what numbers are on the RHS. No factoring or guess work needed
x^2 - y^2 = 24 and xy = 35 =>
(x^2 + y^2)^2 = (x^2 - y^2)^2 + 4x^2y^2 = 24^2 + 4*35^2 = 5476 =>
x^2 + y^2 = 74 =>
(x + y)^2 = x^2 + y^2 + 2xy = 74 + 2*35 = 144 =>
x + y = 12 or -12
From the equation
[(x+y)^2-2xy]^2=(x^2-y^2)^2+(2xy)^2
[(x+y)^2-70]^2=24^2+70^2
so we get
x+y=+-12 , x+y=+-2i
X^2*Y^2=35^2
X^2=Y^2+24
Y=5
X=7
X+Y=12
X^2+Y^2 being the sum of 2 squares is automatically positive.
x(2)=24+y(2)=24+(35/x)(2) -> x(4)=24x(2)+35(2) -> x(4)-2*12x(2)+12(2)=35(2)+12(2) -> [x(2)-12](2) = 37(2) -> x(2)-12 = 37 -> x(2) = 49 -> x=7 -> y=5 -> x+y=12
49-25 = 24 QED (did it in my head in 30 sec.
（x+y）（x-y）=24*1=12*2=8*3=6*4, 考慮xy=35, 故x=7,y=5
You make it so complicated, it is so simple.. It would take 20 seconds to solve, but I don't know what you think
How come a square results in a negative number? X2 + y2 should both be positive numbers, they cannot yield -74!
আপনাকে কী বলবো বলার ভাষা হারিয়ে ফেলেছি
Легко!
x²-y²=24
xy=35
35=7×5
5²-7²=-24
7²-5²=49-25=24
x=7, y=5.
Jika x dan y adalah bilangan riil, maka himpunan penyelesaian dari sistem persamaan tersebut = { ( 7,5 ), ( -7, -5 ) }
( x² - 49 )( x² + 25 ) = 0
He he iya, aku aja bs langsung pastikan 7 dan 5 tanpa bersusah payah
@@Enzo_Himawan3155
x1+y1=12 ✅
x2+y2=-12 ✅
x3+y3=2i ✅
x4+y4=-2i ✅
Let z=x+iy, then z^2=(x^2-y^2)+2i(xy)=24+70i so z =+-z0 where z0=7+5i is a racine of 24+70i which can be found by taking the polar form of 24+70i. then x+y=+-12 and that's all!!!
This maths is so easy! It is very popular at public high school in Vietnam
But x2+y2 always > or =0, that means x2+y2 =74 is true. We dont have check x2+y2=-74.
tell me is it correct or not,
Here xy = 35
So,
X = 35/y
Now puting the value of y in x^2 - y^2 or (x+y) (x-y) = 24
Now,
(35/y + y) ( 35/y -y) = 24
Or, [(35+y^2)/y] [(35-y^2)/y] = 24
Now getting LCM
[(35+y^2)(35-y^2)]/y^2 = 24
Or, (35^2 - y^2)/y^2 = 24
Or, 1225 - y^2 = 24y^2
Or, 1225 = 24y^2 + y^2
Or, 1225 = 25y^2
Or, 1225/25 = y^2
Or, 49 = y^2
So, y = 7
puting the value of y=7 in xy = 35
Now,
x7 = 35
So, x = 35/7 = 5
Now,
(x+y) = (5+7) = 12
Can't we do it
X+y=12
X7 y5
Xy35
X^2-y^2=(x+y)( X--- y)
=(7+5) (7-5)= 35
i. e x+y=12
24 = X2 -- y2 = [ 7× 7 ] -- [ 5×5]
35 = X × y = [ 7×5]
X+ y = 7+5= 12
=> x^2-y^2 = 24
=> xy = 35
Why i guess about 7 and 5 in 20 second or less??
Well not a person really good at maths but
If xy=35, then either x is 7or 5 or -7 or -5 cause 35 is only divisible by these two , so value of x+y is -12 or 12
Great🎉
Thanks ❤️👍🤝