Find the range of values of x for which (2x^{2} + 7x – 15 > 0).
-
A.
(x 5) -
B.
(x frac{3}{2}) -
C.
(-frac{3}{2} -
D.
(-5
Correct Answer: Option B
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)