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

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

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Answer 1 · 2017-01-06T11:16:57

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

Explanation:

We start by factorising the expression

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

and

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

Now, we can make the sign chart

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

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

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

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

Therefore,

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

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