How do you solve $x^2-4x-32>0$ using a sign chart?

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Answer 1 · 2016-11-12T13:56:12

The answer is $x in] -oo,-4 [ uu ] 8, oo[$

Let's factorise the expression $f(x)=x^2-4x-32=(x-8)(x+4)$

let's do the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaa)$ $8$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+4$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x-8$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore, $f(x)>0$ when $x in] -oo,-4 [ uu ] 8, oo[$

graph{x^2-4x-32 [-74, 74.05, -37, 37.14]}

How do you solve x^2-4x-32>0x^2-4x-32>0 using a sign chart?