How do you solve $x^2-3x-4>=0$ using a sign chart?

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How do you solve $x^2-3x-4>=0$ using a sign chart?

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Answer 1 · 2016-11-01T09:17:11

The answer is $-oo < x <=-1$ and $4<= x<+oo$

Explanation:

Start by factorising
$x^2-3x-4=(x+1)(x-4)$
let $y=(x+1)(x-4)$
So we can make the sign chart
$x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-1$ $color(white)(aaaa)$ $4$ $color(white)(aaaa)$ $+oo$
$x+1$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $+$
$x-4$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$
$y$ $color(white)(aaaaaaaa)$ $+$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$

So $x^2-3x-4>=0$
when $-oo < x <=-1$ and when $4<= x<+oo$

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How do you solve $x^2-3x-4>=0$ using a sign chart?

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How do you solve x^2-3x-4>=0x^2-3x-4>=0 using a sign chart?

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How do you solve $x^2-3x-4>=0$ using a sign chart?