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

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

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Answer 1 · 2017-01-15T08:22:49

The answer is $x \in] - 3, 2 [$ $x in ] -3,2 [$

Explanation:

Let's factorise the expression,

$x^{2} + x - 6 = (x - 2) (x + 3)$ $x^2+x-6=(x-2)(x+3)$

and let $f (x) = x^{2} + x - 6$ $f(x)=x^2+x-6$

Now we can do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a a$ $color(white)(aaaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 3$ $x+3$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

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

Therefore,

$f (x) < 0$ $f(x)<0$ , when $x \in] - 3, 2 [$ $x in ] -3,2 [$

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

1 Answer

Explanation:

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How do you solve x^2+x-6<0x2+x−6<0x^2+x-6<0 using a sign chart?

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