Solve the polynomial inequality. express the solution set in in interval notation. $2 x^{2} + 5 x - 3 \leq 0$

(- \infty, - 3) \cup [\frac{1}{2}, \infty]

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[- 3, \frac{1}{2}]

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(- 3, \infty)

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[- \infty, \frac{1}{2}]

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Solution

The correct option is B $[- 3, \frac{1}{2}]$
given, $2 x^{2} + 5 x - 3 \leq 0$
$\Rightarrow 2 x^{2} + 6 x - x - 3 \leq 0$
$\Rightarrow 2 x (x + 3) - 1 (x + 3) \leq 0$
$\Rightarrow (2 x - 1) (x + 3) \leq 0$
$\Rightarrow x \in [- 3, \frac{1}{2}]$
Option B is correct.

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