Carcass wrote:
Which of the following is equivalent to \(\frac{2x^2+8x−24}{2x^2+20x−48}\) for all values of x for which both expressions are defined?
A. \(\frac{x−2}{x−4}\)
B. \(\frac{x−2}{x+4}\)
C. \(\frac{x+2}{x+4}\)
D. \(\frac{x+2}{x−12}\)
E. \(\frac{x+6}{x+12}\)
APPROACH #1: FactoringGiven: \(\frac{2x^2+8x−24}{2x^2+20x−48}\)
Factor the greatest common factor from numerator and denominator: \(\frac{2(x^2+4x−12)}{2(x^2+10x−24)}\)
Factor each quadratic: \(\frac{2(x +6)(x - 2)}{2(x + 12)(x - 2)}\)
Simplify: \(\frac{x +6}{x + 12}\)
Answer: E
APPROACH #2: Testing a valueTwo expressions are equivalent if they evaluate to the same number for all values of x.
For example, we know that 6x + 4x is equivalent to 10x because 6x + 4x = 10x for all values of x.
So, if x = 1.1, we see that 6x + 4x = 6(1.1) + 4(1.1) = 6.6 + 4.4 = 11
Likewise, if x = 1.1, we see that 10x = 10(1.1) = 11 Let's apply this property to the given expression. We'll use a nice number to work with such as \(x = 1\)
We get: \(\frac{2x^2+8x−24}{2x^2+20x−48} = \frac{2(1)^2+8(1)−24}{2(1)^2+20(1)−48} = \frac{-14}{-26} = \frac{7}{13}\)
This tells us that the correct answer must also evaluate to be \(\frac{7}{13}\) when \(x=1\).
Let's plug x=
1 and we chance of choice to see what we get....
A. (
1−2)/(
1−4) = 1/3. No good. We want 7/13. Eliminate.
B. (
1−2)/(
1+4) = -1/5. No good. We want 7/13. Eliminate.
C. (
1+2)/(
1+4) = 3/5. No good. We want 7/13. Eliminate.
D. (
1+2)/(
1−12) = -3/11. No good. We want 7/13. Eliminate.
E. (
1+6)/(
1+12) = 7/13
PERFECTAnswer: E