Question

3x² - 13x + 8 < 4
x² + 6x ≤ 0

Explanation:

Step1: Solve first inequality: rearrange

$3x^2 -13x +8 < 4$
$3x^2 -13x +4 < 0$

Step2: Factor the quadratic

$(3x-1)(x-4) < 0$

Step3: Find critical points

$3x-1=0 \implies x=\frac{1}{3} \approx 0.33$; $x-4=0 \implies x=4$

Step4: Test intervals for sign

Quadrola opens upward, so it is negative between roots: $\frac{1}{3} < x < 4$
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Step5: Solve second inequality: rearrange

$x^2 +6x \leq 0$

Step6: Factor the quadratic

$x(x+6) \leq 0$

Step7: Find critical points

$x=0$; $x+6=0 \implies x=-6$

Step8: Test intervals for sign

Quadrola opens upward, so it is non-positive between roots: $-6 \leq x \leq 0$

Answer:

1. For $3x^2 -13x +8 < 4$: $\boldsymbol{\frac{1}{3} < x < 4}$
2. For $x^2 +6x \leq 0$: $\boldsymbol{-6 \leq x \leq 0}$