Question

Solve the following inequality:
$\frac{3x^2+6x-15}{(2x-1)(x+3)}\ge 1$

• A. (-∞,-4] ∪ (-3,½) ∪ [4,∞)
• B. (-∞,-5] ∪ (-3,½) ∪ [3,∞)
• C. (-∞,-4] ∪ (-2,½) ∪ [3,∞)
• D. (-∞,-4] ∪ (-3,½) ∪ [3,∞)

Answer 1

The correct option is D (-∞,-4] ∪ (-3,½) ∪ [3,∞)

$\frac{3x^2+6x-15}{(2x-1)(x+3)}\ge 1$

$\Rightarrow \frac{3x^2+6x-15}{2x^2+6x-x-3}-1\ge 0$

$\Rightarrow \frac{3x^2+6x-15-(2x^2+6x-x-3)\ge 0}{(2x-1)(x+3)}$

$\Rightarrow \frac{3x^2+6x-15-2x^2+5x-3\ge 0}{(2x-1)(x+3)}$

$\Rightarrow \frac{x^2+x-12\ge 0}{(2x-1)(x+3)}$

$\Rightarrow \frac{x^2+4x-3x-12}{(2x-1)(x+3)}\ge 0$

$\Rightarrow \frac{x(x+4)-3(x+4)}{(2x-1)(x+3)}\ge 0$

​​​​​​​​​​​​​​$\Rightarrow \frac{(x+4)(x-3)}{(2x-1)(x+3)}\ge 0$

​​​​​​​​​​​​​​$\Rightarrow \frac{(x+4)(x-3)(2x-1)(x+3)}{(2x-1{)}^2(x+3{)}^2}\ge 0$
$x\ne \frac{1}{2},x\ne -3$

​​​​​The critical points are $-4,-3,\frac{1}{2}$ and 3.
Also $x\ne \frac{1}{2}$ and $x\ne -3$ as denominator cannot be equal to zero.
​​​​​​​​​​​​​​
Thus the solution set is
(-∞,-4] ∪ (-3,½) ∪ [3,∞)
​​​​​​​So, option d is correct.