How do you solve $(x - 7) (x + 3) \leq 0$ $(x-7)(x+3)<=0$ using a sign chart?

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How do you solve $(x - 7) (x + 3) \leq 0$ $(x-7)(x+3)<=0$ using a sign chart?

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Answer 1 · 2017-01-31T07:16:30

The answer is $x \in [- 3, 7]$ $x in [ -3, 7 ]$

Explanation:

Let $f (x) = (x - 7) (x + 3)$ $f(x)=(x-7)(x+3)$

Let's build the sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a$ $color(white)(aaaa)$ $- 3$ $-3$ $a a a a$ $color(white)(aaaa)$ $7$ $7$ $a a a a$ $color(white)(aaaa)$ $+ \infty$ $+oo$

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

$a a a a$ $color(white)(aaaa)$ $x - 7$ $x-7$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $-$ $-$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

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

Therefore,

$f (x) \leq 0$ $f(x)<=0$ when $x \in [- 3, 7]$ $x in [ -3, 7 ]$

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How do you solve $(x - 7) (x + 3) \leq 0$ $(x-7)(x+3)<=0$ using a sign chart?

1 Answer

Explanation:

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How do you solve $(x - 7) (x + 3) \leq 0$ $(x-7)(x+3)<=0$ using a sign chart?