How do you solve $x^{2} + 6 x - 16 > 0$ $x^2+6x-16>0$ using a sign chart?

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How do you solve $x^{2} + 6 x - 16 > 0$ $x^2+6x-16>0$ using a sign chart?

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Answer 1 · 2016-12-24T14:38:40

The answer is $x \in] - \infty, - 8 [\cup] 2, + \infty [$ $x in ] -oo,-8 [ uu ]2, +oo[$

Explanation:

Let's factorise the expression

$x^{2} + 6 x - 16 = (x + 8) (x - 2)$ $x^2+6x-16=(x+8)(x-2)$

and let $f (x) = (x + 8) (x - 2)$ $f(x)=(x+8)(x-2)$

Now we can do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 8$ $-8$ $a a a a$ $color(white)(aaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 8$ $x+8$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 2$ $x-2$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

Therefore,

$f (x) > 0$ $f(x)>0$ when $x \in] - \infty, - 8 [\cup] 2, + \infty [$ $x in ] -oo,-8 [ uu ]2, +oo[$

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How do you solve $x^{2} + 6 x - 16 > 0$ $x^2+6x-16>0$ using a sign chart?

1 Answer

Explanation:

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