How do you solve the inequality $x^2-4x<=5$ ?

Question

4050 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-01-28T12:29:02

The answer is $x in [ -1, 5 ]$

Let's rewrite the inequality

$x^2-4x-5<=0$
Let's factorise the LHS

$(x+1)(x-5)<=0$

Let $f(x)=(x+1)(x-5)$

Let's build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-1$ $color(white)(aaaaa)$ $5$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $x+1$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

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

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

Therefore,

$f(x)<=0$ when $x in [ -1, 5 ]$

How do you solve the inequality x^2-4x<=5x^2-4x<=5?