How do you solve $x^2-3x>=10$ using a sign chart?

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Answer 1 · 2016-12-11T09:44:03

The answer is $x in ] -oo,-2 ] uu [5, +oo[$

Let's rewrite the equation

$x^2-3x-10>=0$

Let $f(x)=x^2-3x-10=(x+2)(x-5)$

The domain of $f(x)$ is $D_f(x)=RR$

Let's do the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-2$ $color(white)(aaaa)$ $5$ $color(white)(aaaa)$ $+oo$

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

$color(white)(aaaa)$ $x-5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

So,

$f(x)>=0$ when $x in ] -oo,-2 ] uu [5, +oo[$

How do you solve x^2-3x>=10x^2-3x>=10 using a sign chart?