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

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Answer 1 · 2017-07-14T08:39:40

The solution is $x in (-oo, (1-sqrt5)) uu((1+sqrt5), +oo)$

We must first find the roots of the quadratic equations

$x^2-2x-4=0$

The discriminant is

$Delta=b^2-4ac=(-2)^2-4*(1)*(-4)=20$

As $Delta>0$ , there are $2$ real roots

The roots are

$x=(-b+-sqrtDelta)/(2a)$

$x_1=(2-sqrt20)/2=1-sqrt5$

$x_2=(2+sqrt20)/2=1+sqrt5$

Let $f(x)=x^2-2x-4$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaaa)$ $x_1$ $color(white)(aaaa)$ $x_2$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $(x-x_1)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $(x-x_2)$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore,

$f(x)>0$ when $x in (-oo, (1-sqrt5)) uu((1+sqrt5), +oo)$

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