Evaluate \(\frac{x^2 + x - 2}{2x^2 + x -3}\) when x = -1
A
-2
B
-1
C
\(-\frac{1}{2}\)
D
1
Verified Correct Answer: D.
Step-by-Step Mathematical Solution
\(\frac{x^2 + x - 2}{2x^2 + x - 3}\)
= \(\frac{x^2 + 2x - x - 2}{2x^2 + 3x - 2x - 3}\)
= \(\frac{x(x + 2) - 1(x + 2)}{x(2x + 3) - 1(2x + 3)}\)
= \(\frac{(x - 1)(x + 2)}{(x - 1)(2x + 3)}\)
= \(\frac{x + 2}{2x + 3}\)
At x = -1,
= \(\frac{-1 + 2}{2(-1) + 3}\)
= \(\frac{1}{1}\)
= 1
Source: Myschool Past Questions Archive (Record #26485)