How do you solve $16x^2-81>=0$ ?

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Answer 1 · 2017-03-01T15:08:42

The solution is $x in ]-oo,-9/4]uu[9/4,+oo[$

We need

$a^2-b^2=(a+b)(a-b)$

Let's factorise the inequality

$16x^2-81=(4x+9)(4x-9)$

Let $f(x)=(4x+9)(4x-9)$

We build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-9/4$ $color(white)(aaaa)$ $9/4$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $4x+9$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $4x-9$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore,

$f(x)>=0$ when $x in ]-oo,-9/4]uu[9/4,+oo[$

How do you solve 16x^2-81>=016x^2-81>=0?