Evaluate \(\frac{x^2 + x - 2}{2x^2 + x -3}\) when x = -1
MATHEMATICS
WAEC 2001
Evaluate \(\frac{x^2 + x - 2}{2x^2 + x -3}\) when x = -1
- A. -2
- B. -1
- C. \(-\frac{1}{2}\)
- D. 1
Correct Answer: D. 1
Explanation
\(\frac{x^2 + x - 2}{2x^2 + x - 3}\)
= \(\frac{x^2 + 2x - x - 2}{2x^2 + 3x - 2x - 3}\)
= \(\frac{x(x + 2) - 1(x + 2)}{x(2x + 3) - 1(2x + 3)}\)
= \(\frac{(x - 1)(x + 2)}{(x - 1)(2x + 3)}\)
= \(\frac{x + 2}{2x + 3}\)
At x = -1,
= \(\frac{-1 + 2}{2(-1) + 3}\)
= \(\frac{1}{1}\)
= 1
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