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Hi! [waves] i just took the PSAT and this math problem has been bugging me for a looooong time...

(X+Y)(X^2-Y2)=0
In accordance to this equation, which of the following must be true

x=y
x=-y
X^2=Y^2
X^2=-Y^2
X^3=Y^3

2006-10-23 12:46:54 · 7 answers · asked by Aurora 1 in Science & Mathematics Mathematics

7 answers

When you have (x + y)(x² + y²) = 0, that means either (x + y) is zero, or (x² + y²) is zero.

Factoring the right side:
(x + y)(x - y)

So ways to make this equation true are:
x = y
x = -y

For example, x = 5 and y = -5 is one solution, and x = 5 and y = 5 is another valid solution. That means the first two answers *may* be true but aren't necessarily true. The fourth answer *may* be true (only with imaginary numbers or 0). The fifth answer *may* be true, with certain numbers (e.g. x = y).

That only leaves one answer that *MUST* be true:
x² = y²

Very tricky question and it is very easy to get misled into thinking any of the other answers is correct.

Another way to look at this is to factor the equation. That results in (x + y)(x + y)(x - y) = 0

This is equivalent to saying:
(x + y) = 0 *or* (x + y) = 0 *or* (x - y) = 0

Logically this is equivalent to just saying:
(x + y) = 0 or (x - y) = 0, there is no need to repeat (x + y) = 0.

Now put this back into an equation and you get:
(x + y)(x - y) = 0

Expand out:
(x² - y²) = 0

Add y² to both sides and you get:
x² = y²

So again you get answer #3:

2006-10-23 12:56:49 · answer #1 · answered by Puzzling 7 · 0 0

(X+Y)(X^2-Y^2)=0
From the above equation, we can deduce that either
1) X+Y=0, or
2)X^2-Y^2=0

1)X+Y=0
X=-Y
the second option is correct.
It also satisfies X^2-Y^2=0
X=-Y
(-Y)^2-Y^2=0
Y^2-Y^2=0
0=0
LHS = RHS
So there!
Options 1 and 3 satisfy X^2-Y^2=0
But option 2 is the best, as it satisfies both X+Y=0 and X^2-Y^2=0

2006-10-24 11:46:34 · answer #2 · answered by Akilesh - Internet Undertaker 7 · 0 0

x^2 = y^2 .

Try the x,y pairs 1,1 and 1,-1. You will see that all the answers fail one of these two tests except the third answer.

2006-10-23 20:08:34 · answer #3 · answered by M W 1 · 0 0

You can tell by the first part (x + y) that one solution for the answer would be where x = -y. You don't need to mess with the other part.

So the second answer is the correct answer.

2006-10-23 19:53:42 · answer #4 · answered by Alan Turing 5 · 0 0

all except x^2=-y^2 must be true

2006-10-23 19:51:57 · answer #5 · answered by raj 7 · 0 0

x= -y

2006-10-23 19:53:55 · answer #6 · answered by pppjd 1 · 0 0

x=-y

2006-10-23 19:48:42 · answer #7 · answered by travislizzie 2 · 0 0

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