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

(- 3, \frac{1}{2})

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

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

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

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Solution

The correct option is C $[- 3, \frac{1}{2}]$
Given,
$2 x^{2} + 5 x - 3 \leq 0$
$\Rightarrow 2 x^{2} + 6 x - x - 3 \leq 0$
$\Rightarrow 2 x (x + 3) - 1 (x + 3) \leq 0$
$\Rightarrow (2 x - 1) (x + 3) \leq 0$ ... $(i)$
Two cases arises:
Case I:
$(i)$ holds,
if $(2 x - 1) \geq 0$ and $(x + 3) \leq 0$
if $x \geq \frac{1}{2}$ and $x \leq - 3$
No such $x$ exist.
Case II:
$(i)$ holds,
if $(2 x - 1) \leq 0$ and $(x + 3) \geq 0$
if $x \leq \frac{1}{2}$ and $x \geq - 3$
if $- 3 \leq x \leq \frac{1}{2}$
if $x \in [- 3, \frac{1}{2}]$
Option $C$ is correct.

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