How do you solve $x^{2} + 4 x + 4 \geq 9$ $x^2+4x+4>=9$ using a sign chart?

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How do you solve $x^{2} + 4 x + 4 \geq 9$ $x^2+4x+4>=9$ using a sign chart?

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Answer 1 · 2016-10-23T14:27:59

the answer is $x \in (- \infty, - 5) \cup (1, + \infty)$ $x in(-oo,-5)uu(1,+oo)$

Explanation:

$x^{2} + 4 x + 4 \geq 9$ $x^2+4x+4>=9$
so $x^{2} + 4 x - 5 \geq 0$ $x^2+4x-5>=0$
Factorising $(x - 1) (x + 5) \geq 0$ $(x-1)(x+5)>=0$
The values we are looking at are $x = 1$ $x=1$ and $x = - 5$ $x=-5$

the sign chart is the following

$x$ $x$ $a a a a a a a a a a a$ $color(white)(aaaaaaaaaaa)$ $- \infty$ $-oo$ $a a a a a a$ $color(white)(aaaaaa)$ $- 5$ $-5$ $a a a a a a$ $color(white)(aaaaaa)$ $1$ $1$ $a a a a a a$ $color(white)(aaaaaa)$ $+ \infty$ $+oo$
$x - 1$ $x-1$ $a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$
$x + 5$ $x+5$ $a a a a a a a a a a a a a$ $color(white)(aaaaaaaaaaaaa)$ $-$ $-$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$
$(x - 1) (x + 5)$ $(x-1)(x+5)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a a$ $color(white)(aaaaa)$ $+$ $+$

From the sign chart, we see that $x \in (- \infty, - 5)$ $x in (-oo,-5)$ for the function to be positive
From the sign chart, we see that $x \in (1, + \infty)$ $x in (1,+oo)$ for the function to be positive

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How do you solve $x^{2} + 4 x + 4 \geq 9$ $x^2+4x+4>=9$ using a sign chart?

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Explanation:

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How do you solve $x^{2} + 4 x + 4 \geq 9$ $x^2+4x+4>=9$ using a sign chart?