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If x^2 + xy - 32 = 0, and x and y are integers, then y . . .

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by EconomistGMATTutor » Fri Dec 08, 2017 12:44 pm
Hello Vjesus12.

Let's take a look at your question.

If $$x^2+xy-32=0$$ then we could rewrite this as follows: $$x^2+xy-32=0\ \leftrightarrow\ \left(x-a\right)\left(x-b\right)=0$$ where "a" and "b" satisfy $$a+b=y\ \ \ \ \ and\ \ \ \ \ \ a\cdot b=-32.$$ The possible values for "a" and "b" are: $$\left(a,b\right)=(1,-32);\ (2,-16);\ (4,-8);(-1,32);(-2,16);(-4,8).$$ So, the possible sums a+b are: $$-31,\ -14,\ \ -4,\ \ 31,\ \ 14,\ \ \ 4.$$ So, the correct answer is option C.

I hope this explanation can help you.

I'm available if you'd like a follow up.

Regards.
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by GMATWisdom » Fri Dec 08, 2017 1:18 pm
VJesus12 wrote:If x^2 + xy - 32 = 0, and x and y are integers, then y could equal each of the following except?

A) -31
B) -14
C) 2
D) 4
E) 14

The OA is C.

How can I find the value of y? Should I use the quadratic equation? I am confused.
y should be the sum of two factors of -32

factors could be
1. -16 and 2 their sum is -14
2. -8 and 4 ............ -4
3. -4 and 8 ............. +4
4. -16 and 2 ............. -14
5 -1 and 32............. +31
6 -32 and 1...............-31

thus there are no two factors of -32 whose sum is 2

hence C is the correct answer.
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