If x^2 + y^2 = 29, what is the value of (x-y)^2?

[HarrisonI74]

If x^2 + y^2 = 29, what is the value of (x-y)^2?

(1) xy = 10
(2) x = 5


Can anyone explain why statement 2 is insufficient if I can plug in the value of X in the same equation used to solve statement 1?

(x-y)^2 => x^2 - 2xy + y^2
And, x^2 + y^2 = 29 => 29 - 2xy

if x = 5 then 29- 2(5)y = 29 - 10y
-10y = 29
y = 29/-10
y = -2.9

[Sage Pearce-Higgins]

You're not quite clear with your algebra, specifically, what's an equation and what's a question.

(x-y)^2 => x^2 - 2xy + y^2

Good, rephrase the question. However, I don't know what "=>" means here; it should be "=".
So, the new question is "What's the value of x^2 - 2xy + y^2?"

And, x^2 + y^2 = 29 => 29 - 2xy

This seems confused, is it an equation, or a question? I'd recommend that you think (and write):
x^2 + y^2 = 29, so the question get's rephrased as "What's the value of 29 - 2xy ?"
At this point, it should be clear that statement 1 is sufficient, as it gives you the missing unknowns in the question as values.

if x = 5 then 29- 2(5)y = 29 - 10y
-10y = 29

This is your real mistake. Where did this equation come from? I guess that you've assumed that 29 - xy = 0, but that's stated nowhere in the problem. You've made an equation out of the question. Be sure that your scratch work helps you avoid this kind of confusion (which is the big trap of problems like this one).

[PIYUSHT767]

Hey, I think I got your doubt.It's a pretty common doubt to have.


When you plug in the value of x = 5 in the question's equation, you're assuming that you'll be able to find the right value of y (which is 2, in this case).

However, don't forget to consider the negative value of y which could be possible while taking the square root.

Hence, the value of y could be +2 or -2, if you plug in the value of x=5 in the equation given in the question.


Hope this helps!

[Sage Pearce-Higgins]

I agree, PIYUSHT767. That's a big trap for this question.