How do you solve $x^{2} - 2 x - 5 \leq 0$ $x^2-2x-5<=0$ using a sign chart?

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How do you solve $x^{2} - 2 x - 5 \leq 0$ $x^2-2x-5<=0$ using a sign chart?

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Answer 1 · 2016-10-30T19:25:37

The answer is $1 - \sqrt{6} \leq x \leq 1 + \sqrt{6}$ $1-sqrt6<=x<=1+sqrt6$

Explanation:

First we start by solving $y = x^{2} - 2 x - 5 = 0$ $y=x^2-2x-5=0$
We calculate $Delta=b^2-4ac=4+20=24$

So the roots are $=(2+-sqrt24)/2=(2+-2sqrt6)/2$
so the roots are $1+sqrt6$ and $1-sqrt6$
We can do the sign chart

$x$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaa)$ $1-sqrt6$ $color(white)(aaa)$ $1+sqrt6$ $color(white)(aaa)$ $+oo$
$1-sqrt6$ $color(white)(aaaa)$ $-$ $color(white)(aaa)$ $0$ $color(white)(aaaa)$ $+$ $color(white)(aaaaa)$ $+$
$1+sqrt6$ $color(white)(aaaa)$ $-$ $color(white)(aaa)$ #color(white)(aaaaa)-# $color(white)(aaaaa)$ $+$
$y$ $color(white)(aaaaaaaaa)$ $+$ $color(white)(aaaaaaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaa)$

So $x^2-2x-5<=0$ when $1-sqrt6<=x<=1+sqrt6$

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How do you solve $x^{2} - 2 x - 5 \leq 0$ $x^2-2x-5<=0$ using a sign chart?

1 Answer

Explanation:

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How do you solve $x^{2} - 2 x - 5 \leq 0$ $x^2-2x-5<=0$ using a sign chart?