How do you solve $(x + 5) (x - 2) \geq 0$ $(x+5)(x-2)>=0$ ?

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How do you solve $(x + 5) (x - 2) \geq 0$ $(x+5)(x-2)>=0$ ?

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Answer 1 · 2016-12-20T15:40:20

The answer is $x \in] - \infty, - 5] \cup [2, + \infty [$ $x in ] -oo,-5 ] uu [2, +oo[$

Explanation:

Let $f (x) = (x + 5) (x - 2)$ $f(x)=(x+5)(x-2)$

Now, we can do a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 5$ $-5$ $a a a a$ $color(white)(aaaa)$ $2$ $2$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 5$ $x+5$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

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

Therefore,

$f (x) \geq 0$ $f(x)>=0$ , when $x \in] - \infty, - 5] \cup [2, + \infty [$ $x in ] -oo,-5 ] uu [2, +oo[$

graph{(x+5)(x-2) [-28.86, 28.85, -14.43, 14.43]}

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