How do you solve the inequality $\frac{x^{2} - 16}{x^{2} - 4 x - 5} \geq 0$ $(x^2-16)/(x^2-4x-5)>=0$ ?

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How do you solve the inequality $\frac{x^{2} - 16}{x^{2} - 4 x - 5} \geq 0$ $(x^2-16)/(x^2-4x-5)>=0$ ?

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Answer 1 · 2018-04-16T09:28:16

The solution is $x \in (- \infty, - 4] \cup (- 1, 4] \cup (5, + \infty)$ $x in (-oo,-4] uu(-1,4] uu (5,+oo)$

Explanation:

The numerator is

$x^{2} - 16 = (x - 4) (x + 4)$ $x^2-16=(x-4)(x+4)$

The denominator is

$x^{2} - 4 x - 5 = (x - 5) (x + 1)$ $x^2-4x-5=(x-5)(x+1)$

Let

$f (x) = \frac{(x - 4) (x + 4)}{(x - 5) (x + 1)}$ $f(x)=((x-4)(x+4))/((x-5)(x+1))$

Let's build a sign chart

$a a a a$ $color(white)(aaaa)$ $x$ $x$ $a a a a$ $color(white)(aaaa)$ $- \infty$ $-oo$ $a a a a a$ $color(white)(aaaaa)$ $- 4$ $-4$ $a a a a$ $color(white)(aaaa)$ $- 1$ $-1$ $a a a a a$ $color(white)(aaaaa)$ $4$ $4$ $a a a a a$ $color(white)(aaaaa)$ $5$ $5$ $a a a a a a$ $color(white)(aaaaaa)$ $+ \infty$ $+oo$

$a a a a$ $color(white)(aaaa)$ $x + 4$ $x+4$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x + 1$ $x+1$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aaaa)-# $a$ $color(white)(a)$ $∣ ∣$ $||$ $a$ $color(white)(a)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 4$ $x-4$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aaaa)-# $a$ $color(white)(a)$ #color(white)(aa)-# $a$ $color(white)(a)$ $0$ $0$ $a a$ $color(white)(aa)$ $+$ $+$ $a a a a$ $color(white)(aaaa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $x - 5$ $x-5$ $a a a a a$ $color(white)(aaaaa)$ $-$ $-$ $a a a$ $color(white)(aaa)$ #color(white)(aaaa)-# $a$ $color(white)(a)$ #color(white)(aa)-# $a$ $color(white)(a)$ #color(white)(aaa)-# $a$ $color(white)(a)$ $∣ ∣$ $||$ $a a$ $color(white)(aa)$ $+$ $+$

$a a a a$ $color(white)(aaaa)$ $f (x)$ $f(x)$ $a a a a a a$ $color(white)(aaaaaa)$ $+$ $+$ $a a a$ $color(white)(aaa)$ $0$ $0$ $a a a$ $color(white)(aaa)$ $-$ $-$ $a$ $color(white)(a)$ $∣ ∣$ $||$ $a$ $color(white)(a)$ $+$ $+$ $a$ $color(white)(a)$ $0$ $0$ $a a$ $color(white)(aa)$ $-$ $-$ $a$ $color(white)(a)$ $||$ $color(white)(aa)$ $+$

graph{((x+4)(x-4))/((x-5)(x+1)) [-14.24, 14.23, -7.12, 7.12]}

Therefore,

$f(x)>=0$ when $x in (-oo,-4] uu(-1,4] uu (5,+oo)$

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How do you solve the inequality $\frac{x^{2} - 16}{x^{2} - 4 x - 5} \geq 0$ $(x^2-16)/(x^2-4x-5)>=0$ ?

1 Answer

Explanation:

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How do you solve the inequality x2−16x2−4x−5≥0(x^2-16)/(x^2-4x-5)>=0?

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How do you solve the inequality $\frac{x^{2} - 16}{x^{2} - 4 x - 5} \geq 0$ $(x^2-16)/(x^2-4x-5)>=0$ ?