How do you solve $5x<=2-3x^2$ using a sign chart?

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Answer 1 · 2017-04-23T06:47:26

The solution is $x in [-2,1/3]$

Let's rearrange and factorise the inequality

$5x<=2-3x^2$

$3x^2+5x-2<=0$

$(3x-1)(x+2)<=0$

Let $f(x)=(3x-1)(x+2)$

We build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-2$ $color(white)(aaaaa)$ $1/3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+2$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $3x-1$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x9$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<=0$ when $x in [-2,1/3]$

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