How do you solve $(a - 2) (a + 4) > 0$ $(a-2)(a+4)>0$ ?

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How do you solve $(a - 2) (a + 4) > 0$ $(a-2)(a+4)>0$ ?

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Answer 1 · 2016-12-05T16:15:12

The answer is $x \in] - \infty, - 4 [\cup] 2, + \infty [$ $x in ] -oo,-4 [ uu ] 2,+oo [$

Explanation:

Let $f (a) = (a - 2) (a + 4)$ $f(a)=(a-2)(a+4)$

Then we do a sign chart

$a a a a$ $color(white)(aaaa)$ $a$ $a$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 4$ $-4$ $a a a a$ $color(white)(aaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $a + 4$ $a+4$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

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

For $f (a) > 0$ $f(a)>0$ , $x \in] - \infty, - 4 [\cup] 2, + \infty [$ $x in ] -oo,-4 [ uu ] 2,+oo [$

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How do you solve $(a - 2) (a + 4) > 0$ $(a-2)(a+4)>0$ ?

1 Answer

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