How do you solve $x^2+x-6>0$ ?

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Answer 1 · 2017-01-06T11:16:57

The answer is $x in ] -oo,-3[ uu ] 2, oo[$

We start by factorising the expression

$x^2+x-6=(x-2)(x+3)$

and

let $f(x)=(x-2)(x+3)$

Now, we can make the sign chart

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

$color(white)(aaaa)$ $x+3$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

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

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

Therefore,

$f(x)>0$ , when $x in ] -oo,-3[ uu ] 2, oo[$

How do you solve x^2+x-6>0x^2+x-6>0?