How do you solve $2x+2>x^2$ using a sign chart?

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Answer 1 · 2017-05-23T05:16:18

The solution is $x in (1-sqrt3, 1+sqrt3)$

Let's rewrite the equation

$2x+2>x^2$

$x^2-2x-2<0$

The roots of the equation

$x^2-2x-2=0$

are $x_1$ and $x_2$

$x_2=(2+sqrt(2^2-(4*1*-2)))/2=(2+sqrt12)/2=(2+2sqrt3)/2=1+sqrt3$

$x_1=(2-sqrt(2^2-(4*1*-2)))/2=(2-sqrt12)/2=(2-2sqrt3)/2=1-sqrt3$

Let $f(x)=x^2-2x-2$

We build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaaa)$ $x_1$ $color(white)(aaaa)$ $x_2$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-x_1$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-x_2$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(x)<0$ , when $x in (x_1,x_2)$

How do you solve 2x+2>x^22x+2>x^2 using a sign chart?