Real Practice Test

[guest]

Does anyone know how you would solve this algebraically?

Is radical((x-3)^2)=3-x?

1. x does not equal 3
2. -x*absolute value(x)>0

Thanks.[/list]

[Guest]

The answer should be (D)

Lets rephrase the question stem

(x-3)^2 = 3-x
x^2 + 9 - 6x = 3-x
x^2 - 5x + 6 = 0
(x-2)(x-3) = 0

So what has really been asked is, whether x = 2,3?

from statement (1) we know x is not 3, so x is 2. Hence sufficient
from statement (2) -x*|x|>0, this condition would only hold for -ve values hence according to this
statement x<0 But the questions asks if x=2,3? So x is -ve hence x is neither 2 nor 3, Hence sufficient.

Hope it helps

GMAT 2007

[guest]

I'm confused as to how we were able to get from the equation given in the question to your rephrased version. How were you able to get rid of the radical around (x-3)^2?

[GMAT 2007]

If you apply the following formula to the radical (x-3)^2 and solve the equation you would get x=2,3
(a-b)^2 = a^2 + b^2 - 2ab

GMAT 2007

[shaji]

Statement1: If x not equal to 3 does not imply that x=2. it could be any any number other than 3.

The answer should be (D)

Lets rephrase the question stem

(x-3)^2 = 3-x
x^2 + 9 - 6x = 3-x
x^2 - 5x + 6 = 0
(x-2)(x-3) = 0

So what has really been asked is, whether x = 2,3?

from statement (1) we know x is not 3, so x is 2. Hence sufficient
from statement (2) -x*|x|>0, this condition would only hold for -ve values hence according to this
statement x<0 But the questions asks if x=2,3? So x is -ve hence x is neither 2 nor 3, Hence sufficient.

Hope it helps

GMAT 2007

[guest]

I feel really dumb....

Even if we were to solve (x-3)^2= x^2-6x+9, isn't that still radical(x^2-6x+9)?

If we were to square both sides to remove the radical then wouldn't the equation be x^2-6x+9=(3-x)^2.

I must be missing something

[GMAT 2007]

I am sorry I misunderstood the problem earlier.
radical((x-3)^2) = x-3 if we rephrase the statement

x-3 = 3-x or 2x = 6 or x=3

Note: We don't have to square the equation, because the left hand side is linear already.

What has really been asked is x =3 ?

According to statement (A) x is not equal to 3, so Sufficient,
Similarly statement (B) says x is -ve, but so x is not equal to 3.

Hence the answer is (D)

[Guest]

Hmmm... the answer is actually B - though I did answer it as D. I noticed that when x<=3, the solution works - which is why statement 1 isn't sufficient. However, does anyone know how to reach that algebraically?

[esledge]

Hey everyone, this is a really tricky one!

The thing to remember is that when you take the square root of a plain-old number, not an expression, the GMAT always means the positive root. For example, sqrt(3^2) = sqrt(9) = 3, NOT -3. Similarly, sqrt[(-3)^2] = sqrt[9] = 3, NOT -3, even though -3 was the original base we squared inside the radical. More generally, sqrt(x^2) = |x|.

So we rephrase the question this way:
sqrt[(x - 3)^2]=(3 - x)?
|x - 3| = (3 - x)?

Since the expression in the absolute value signs could be positive, zero or negative, we must account for two cases:
Non-Negative Case (when x - 3 >= 0, or x >= 3)
(x - 3) = (3 - x)?
2x = 6?
x = 3?
The answer will be a definite "yes" when x = 3, but "no" when x > 3.

Negative Case (when x - 3 < 0, or x < 3)
-(x - 3) = (3 - x)?
-x + 3 = (3 - x)?
0 = 0?
Obviously, zero is always equal to zero, so we get a definite "yes" answer when x < 3.

Putting these cases together, we get a "yes" answer when x =<3, but a "no" answer when x > 3.

Our rephrased question is thus: "Is x less than or equal to 3?"

(1) INSUFFICIENT: If x does not equal 3, x could be less than 3 ("yes" answer) or greater than 3 ("no" answer).

(2) SUFFICIENT: -x*|x|>0. First, note that x cannot equal zero, since that would make -x*|x| = 0. The absolute value of x must be positive, no matter what the sign of x is. The only way for -x*|x| = (-x*positive) to be positive is for -x > 0. Thus, x < 0, so the answer is "yes."

The correct answer is B.