How do you solve $x^2-1>=4x$ using a sign chart?

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Answer 1 · 2016-12-14T08:33:29

The answer is $x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[$

Let's rewrite the equation

$x^2-4x-1>=0$

Let $f(x)=x^2-4x-1$

We need the values of $x$ , when $f(x)=0$

that is, $x^2-4x-1=0$

Let's calculate $Delta=16-4*1*-1=20$

$Delta>0$ . there are 2 real solutions

$x_1=(4+sqrt20)/2=(4+2sqrt5)/2=2+sqrt5$

$x_2=(4-sqrt20)/2=(4-2sqrt5)/2=2-sqrt5$

Now, we can do our sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaaa)$ $-oo$ $color(white)(aaaa)$ $(2-sqrt5)$ $color(white)(aaaa)$ $(2+sqrt5)$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x-2+sqrt5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaaaaaaa)$ $+$ $color(white)(aaaaaaaaa)$ $+$

$color(white)(aaaa)$ $x-2-sqrt5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaaaaaaa)$ $-$ $color(white)(aaaaaaaaa)$ $+$

$color(white)(aaaa)$ $f(x)$ $color(white)(aaaaaaaaaaa)$ $+$ $color(white)(aaaaaaaaaa)$ $-$ $color(white)(aaaaaaaaa)$ $+$

Therefore,

$f(x>=0)$ when $x in ] -oo,(2-sqrt5) ] uu [(2+sqrt5), +oo[$

graph{x^2-4x-1 [-11.25, 11.25, -5.625, 5.625]}

How do you solve x^2-1>=4xx^2-1>=4x using a sign chart?