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jam.hui
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Exponential Equations

by jam.hui Sat Oct 08, 2011 1:49 am

Hi everyone, I need some help with a question that was on a MGMAT CAT. I understand the answer explanation, but want to know why I cannot use another method to solve. Please help.

Question: rad(xy) = xy, what is x+y?

1) x= -1/2
2) y is not equal to 0

The answer is C with the explanation below:

Let’s start by rephrasing the question. If we square both sides of the equation we get:


Now subtract xy from both sides and factor:
(xy)2 - xy = 0
xy(xy - 1) = 0
So xy = 0 or 1

To find the value of x + y here, we need to solve for both x and y.
If xy = 0, either x or y (or both) must be zero.
If xy = 1, x and y are reciprocals of one another.
While we can’t come up with a precise rephrasing here, the algebra we have done will help us see the usefulness of the statements.

(1) INSUFFICIENT: Knowing that x = -1/2 does not tell us if y is 0 (i.e. xy = 0) or if y is -2 (i.e. xy = 1)

(2) INSUFFICIENT: Knowing that y is not equal to zero does not tell us anything about the value of x; x could be zero (to make xy = 0) or any other value (to make xy = 1).

(1) AND (2) SUFFICIENT: If we know that y is not zero and we have a nonzero value for x, neither x nor y is zero; xy therefore must equal 1. If xy = 1, since x = -1/2, y must equal -2. We can use this information to find x + y, -1/2 + (-2) = -5/2.

The correct answer is C.

However, I tried to solve this by plugging and chugging statement one into an updated equation. I squared both sides so the new equation is: xy= (xy)^2. I then plugged the x value from statement 1 and found that A was sufficient. Can someone help me explain why this is not true?

Here is what I did
1. If xy =(xy)^2 and I plug in -1/2 as x into the equation, I get
(-1/2)y= (-1/2)^2 y^2,
2. I then handle the parantheses to show
(-1/2)y= (1/4)y^2.
3. Multiply the entire equation by 4, and get -2y=y^2,
4. divide by y on both sides and get y= -2 and therefore sufficient.

Where did I go wrong here? Please help.

Thanks!
700+
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Re: Exponential Equations

by 700+ Tue Oct 11, 2011 6:41 pm

As per statement 1, x = -1/2. The value of y is not mentioned.

As you calculated, when x = -1/2, y = -2. Another possibility is when x = -1/2, y could be equal to 0. The condition in the question is also satisfied when y = 0.

Therefore, when
x= -1/2 & y = (-2)
x + y = -5/2

When
x = -1/2 & y = 0
x + y = -1/2

Since (x+y) has 2 possible values, A is not sufficient.

Hope I was able to clear your doubt.
RonPurewal
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Re: Exponential Equations

by RonPurewal Sat Oct 15, 2011 2:53 am

700+ Wrote:As per statement 1, x = -1/2. The value of y is not mentioned.

As you calculated, when x = -1/2, y = -2. Another possibility is when x = -1/2, y could be equal to 0. The condition in the question is also satisfied when y = 0.

Therefore, when
x= -1/2 & y = (-2)
x + y = -5/2

When
x = -1/2 & y = 0
x + y = -1/2

Since (x+y) has 2 possible values, A is not sufficient.

Hope I was able to clear your doubt.


nicely done.