How do you solve $5 x \leq 2 - 3 x^{2}$ $5x<=2-3x^2$ ?

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How do you solve $5 x \leq 2 - 3 x^{2}$ $5x<=2-3x^2$ ?

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Answer 1 · 2016-10-16T07:14:59

$- 2 \leq x \leq \frac{1}{3}$ $-2<=x<=1/3$

Explanation:

$5 x \leq 2 - 3 x^{2}$ $5x<=2-3x^2$
Rearranging
$3 x^{2} + 5 x - 2 \leq 0$ $3x^2+5x-2<=0$
$(3 x - 1) (x + 2) \leq 0$ $(3x-1)(x+2)<=0$
$x = \frac{1}{3} or x = - 2$ $x=1/3 or x=-2$
$- 2 \leq x \leq \frac{1}{3}$ $-2<=x<=1/3$
Checking, let $x = - 1$ $x=-1$
$- 4 \cdot 1 = - 4$ $-4*1=-4$ which $\leq 0$ $<=0$

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How do you solve $5 x \leq 2 - 3 x^{2}$ $5x<=2-3x^2$ ?

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