Question

question
what is the solution of the following inequality?
\\(2x^2 + 7x \leq 4\\)

answer
\\(\\{x|x \leq -4\text{ or }x \geq \frac{1}{2}\\}\\) \\(\\{x|x \leq -\frac{1}{2}\text{ or }x \geq 4\\}\\)
\\(\\{x|-\frac{1}{2} \leq x \leq 4\\}\\) \\(\\{x|-4 \leq x \leq \frac{1}{2}\\}\\)

Explanation:

Step1: Rearrange to standard form

$2x^2 + 7x - 4 \leq 0$

Step2: Factor the quadratic

$(2x - 1)(x + 4) \leq 0$

Step3: Find critical points

$2x - 1 = 0 \implies x = \frac{1}{2}$; $x + 4 = 0 \implies x = -4$

Step4: Test intervals for sign

Test $x < -4$: $(-)(-) = + > 0$ (not valid)
Test $-4 < x < \frac{1}{2}$: $(-)(+) = - < 0$ (valid)
Test $x > \frac{1}{2}$: $(+)(+) = + > 0$ (not valid)

Step5: Include critical points

Inequality is $\leq 0$, so include $x=-4$ and $x=\frac{1}{2}$

Answer:

$\{x|-4 \leq x \leq \frac{1}{2}\}$