How do solve $x^{2} + x < 12$ $x^2+x<12$ and write the answer as a inequality and interval notation?

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How do solve $x^{2} + x < 12$ $x^2+x<12$ and write the answer as a inequality and interval notation?

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Answer 1 · 2017-01-16T13:01:31

The answer is $x \in] - 4, 3 [$ $x in ] -4,3 [$
$- 4 < x < 3$ $-4 < x < 3$

Explanation:

The inequation is

$x^{2} + x - 12 < 0$ $x^2+x-12<0$

Let's factorise the expression

$x^{2} + x - 12 = (x - 3) (x + 4)$ $x^2+x-12=(x-3)(x+4)$

Let $f (x) = x^{2} + x - 12$ $f(x)=x^2+x-12$

Now, we can do the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a a$ $color(white)(aaaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 4$ $-4$ $a a a a$ $color(white)(aaaa)$ $3$ $3$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$

$a a a a$ $color(white)(aaaa)$ $x + 4$ $x+4$ $a a a a a a$ $color(white)(aaaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

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

Therefore,

$f (x) < 0$ $f(x)<0$ , when $x \in] - 4, 3 [$ $x in ] -4,3 [$

or $- 4 < x < 3$ $-4 < x <3$

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How do solve $x^{2} + x < 12$ $x^2+x<12$ and write the answer as a inequality and interval notation?

1 Answer

Explanation:

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Impact of this question

Science

Math

Humanities

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How do solve x2+x<12x^2+x<12 and write the answer as a inequality and interval notation?

1 Answer

Explanation:

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How do solve $x^{2} + x < 12$ $x^2+x<12$ and write the answer as a inequality and interval notation?