An algebra problem by Aneesh Kundu

Algebra Level 5

Solve the following system $xy^2=15x^2+17xy+15y^2$

$x^2y=20x^2+3y^2$

$x\not =0, y\not =0$

Enter your answer as the product of all possible distinct complex values of $x$ and $y$ .

The answer is 8867965.

1 solution

Aneesh Kundu
Nov 27, 2014

Divide the 1st eqn by the 2nd one. Cross multiply. Factor out $x^2+y^2$ . Get the complex solutions by substituting $x$ as $\pm iy$ , and the real ones by substituting $y$ as $5x$ .

I will add an elaborate solution in the next 48 hrs.

EDIT

The solution is already below.

Following the steps that you outline, we get $15x^3 - 3x^2 y + 15xy^2 - 3y^3 = 0,$ which factors as $3(5x - y)(x^2 + y^2) = 0,$ so $y = 5x$ , $y = xi$ , or $y = -xi$ .

If $y = 5x$ , then we can substitute into the equation $x^2 y = 20x^2 + 3y^2$ , to get $5x^3 = 20x^2 + 75x^2$ . Then $x = 19$ and $y = 95$ .

If $y = xi$ , then we can substitute into the equation $x^2 y = 20x^2 + 3y^2$ , to get $ix^3 = 17x^2$ . Then $x = -17i$ , and $y = 17$ .

If $y = -xi$ , then we can substitute into the equation $x^2 y = 20x^2 + 3y^2$ , to get $-ix^3 = 17x^2$ . Then $x = 17i$ , and $y = 17$ .

The product of all the solutions is $19 \cdot 95 \cdot (-17i) \cdot 17 \cdot (17i) \cdot 17 = 150,755,405$ .

Jon Haussmann - 6 years, 6 months ago

$y$ can take the values $17, 95$ and $x$ can take the values $19, 17i, -17i$ And so the product should be $17\times 95\times 19\times 17i\times -17i=8867965$

The question clearly said product of all values of x and y, not product of all the ordered pairs

Aneesh Kundu - 6 years, 6 months ago

You're right, my apologies. I should have read the problem more carefully.

Jon Haussmann - 6 years, 6 months ago

@Jon Haussmann – I agree that the question wasn't clearly phrased and could be interpreted differently. Those who answered $8867965 \times 17 = 150,755,405$ have been marked correct.

I've updated the phrasing of the problem.

Calvin Lin Staff - 6 years, 6 months ago

@Calvin Lin – But I got it wrong?? It was my first answer

Kunal Gupta - 6 years, 5 months ago

Shouldn't it be $-17i\times17i=289$ since you are asking for the product of all complex values? @Calvin Lin @Aneesh Kundu

Marc Vince Casimiro - 6 years, 5 months ago

All real numbers are considered complex numbers with their imaginary part 0

Shivam Hinduja - 6 years, 5 months ago

But if the question was asking all solutions, why not just state it? If another question, then, is asking for the product of complex solutions and you entered the product of all(including real ones) and yet wrong - it would be debatable. Hence, complex is complex, it should not be interpreted as real numbers in most cases.

Marc Vince Casimiro - 6 years, 5 months ago

@Marc Vince Casimiro – Whenever a question asks for solutions, it should be assumed that $x$ is real.(unless otherwise stated).

But if a question asks for complex solutions, then both real and non-real solutions should be considered, as real numbers are a subset of complex numbers.

Aneesh Kundu - 6 years, 5 months ago

@Aneesh Kundu – Hm... what if the question needs the complex solutions? What word should I use other than complex such that it would not be mistaken as a real number? I still disagree though...

Marc Vince Casimiro - 6 years, 5 months ago

@Marc Vince Casimiro – When I first learnt about complex numbers, I asked my teacher that if $Im(z)=0$ then its not a complex number?

The teacher replied that every real number is a complex number( even 0 is complex) with imaginary part 0.

If u look up the definition of a complex number, it says

$z$ is a complex number if it can be expressed in the form $a+ib$ where $a$ and $b$ are real numbers and $i=\sqrt{-1}$

You might notice that the definition does not say that $b\not = 0$ .

You can use that the word non-real to describe complex numbers in which $Im(z)\not = 0$ . Note that on this case both $a$ and $b$ can't be simultaneously 0.

Aneesh Kundu - 6 years, 5 months ago

An algebra problem by Aneesh Kundu

The answer is 8867965.

1 solution

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