Explanation

To compare x with –2, you should first solve the equation \(\frac{x(x-2)}{(x+3)(x-4)^2}=0\) for x.

To solve the equation, recall that a fraction \(\frac{a}{b}\) is equal to 0 only if a = 0 and b ≠ 0.

So the fraction \(\frac{x(x-2)}{(x+3)(x-4)^2}=0\) is equal to 0 only if \(x(x - 2)\) is equal to 0 and \((x + 3)(x - 4)^2\) is not equal to 0. Note that the only values of x for which \(x(x - 2)\) = 0 are x = 0 and x = 2, and for these
two values, \((x + 3)(x - 4)^2\) is not equal to 0. Therefore the only values of x for which the fraction \(\frac{x(x - 2)}{(x+3)(x-4)^2}=0\) is equal to 0 are x = 0 and x = 2. Since both 0 and 2 are greater than Quantity B, –2, the correct answer is Choice A.
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x cannot be equal to -2 because the denominator will become 0, that leaves x = 0. Hence A

KEY PROPERTY: If x/y = 0, then we can be certain that x = 0

Aside: Some students will incorrectly state that, if x/y = 0, then y = 0.
However, this assertion is false since x/0 is not an actual number.
We say that x/0 is undefined (it has no value)
So, for example 7/0 is undefined, 23/0 is undefined, and (-5)/0 is undefined

GIVEN: \(\frac{x(x-2)}{(x-3)(x-4)^2}=0\)

This means that x(x-2)= 0
When we solve this equation, we see that EITHER x = 0 OR x = 2

Let's examine each possible case.

case i: x = 0
We get:
QUANTITY A: 0
QUANTITY B: -2
In this case, Quantity A is greater


case ii: x = 2
We get:
QUANTITY A: 2
QUANTITY B: -2
In this case, Quantity A is greater

There are only two possible values of x, and each value is greater than -2
So, we can be certain that Quantity A is greater than Quantity B

Answer: A

Cheers,
Brent
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