Find the range of values of x for which 2x\(^2\) + 7x – 15 ≥ 0.
-
x ≤ -5 or x ≥ \(\frac{3}{2}\)
- x ≥ -5 or x ≤\(\frac{3}{2}\)
- -5 ≤ x ≤ \(\frac{3}{5}\)
- \(\frac{3}{5}\) ≤ x ≤ -5
The correct answer is: A
Explanation
2x\(^2\) + 7x - 15 ≥ 0
2x\(^2\) -3x + 10x - 15 ≥ 0
x(2x - 3) + 5(2x - 3) ≥ 0
(x+5)(2x-3) ≥ 0
the points on x-axis where the graph ≥ 0
x ≤ -5 or x ≥ \(\frac{3}{2}\)