Find the range of values of x for which \(2x^{2} + 7x - 15 >
FURTHER MATHEMATICS
WAEC 2008
Find the range of values of x for which \(2x^{2} + 7x - 15 > 0\).
- A. \(x< -\frac{3}{2}\) or \(x > 5\)
- B. \(x< -5\) or \(x > \frac{3}{2}\)
- C. \(-\frac{3}{2}< x< 5\)
- D. \(-5< x< \frac{3}{2}\)
Correct Answer: B. \(x< -5\) or \(x > \frac{3}{2}\)
Explanation
\(2x^{2} + 7x - 15 > 0 \implies 2x^{2} - 3x + 10x - 15 > 0\)
\(x(2x - 3) + 5(2x - 3) > 0\)
\((x + 5)(2x - 3) > 0\)
For their product to be positive, they are either both +ve or -ve.
\(x + 5 > 0 \implies x > -5\)
\(2x - 3 > 0 \implies 2x > 3\)
\(x > \frac{3}{2}\)
Check:
\(x > -5: x = -3\)
\(2(-3)^{2} + 7(-3) - 15 = 18 - 21 - 15 = -18 < 0\) (Not satisfied)
\(\therefore x < -5\)
\(x > \frac{3}{2}: x = 2\)
\(2(2^{2}) + 7(2) - 15 = 8 + 14 - 15 = 7 > 0\) (Satisfied)
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