How do you solve $1/2x^2<=x+12$ using a sign chart?

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Answer 1 · 2017-08-09T13:31:49

The solution is $x in [-4 ,6]$

Let's rearrange the inequality

$1/2x^2<= x+12$

$1/2x^2- x-12<=0$

$x^2- 2x-24 <=0$

Factorising, we get

$(x+4)(x-6) <=0$

Let, $f(x)=(x+4)(x-6)$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $+oo$ $color(white)(aaaaa)$ $-4$ $color(white)(aaaaaa)$ $6$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+4$ $color(white)(aaaaaa)$ $-$ $color(white)(aa)$ $0$ $color(white)(aaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-6$ $color(white)(aaaaaa)$ $-$ $color(white)(aa)$ $color(white)(aaaa)$ $-$ $color(white)(a)$ $0$ $color(white)(aa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaa)$ $+$ $color(white)(aa)$ $0$ $color(white)(aaa)$ $-$ $color(white)(a)$ $0$ $color(white)(aa)$ $+$

Therefore,

$f(x) <=0$ when $x in [-4 ,6]$

graph{1/2x^2-x-12 [-16.22, 15.8, -12.55, 3.47]}

How do you solve 1/2x^2<=x+121/2x^2<=x+12 using a sign chart?