How do you solve $3(2w^2-5)<w$ using a sign chart?

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Answer 1 · 2017-01-31T18:15:19

The answer is $w in]-3/2, 5/3 [$

Let's rewrite the inequality

$6w^2-w-6<0$

Let's factorise

$(3w-5)(2w+3)<0$

Let $f(w)=(3w-5)(2w+3)$

Now, we can build the sign chart

$color(white)(aaaa)$ $w$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-3/2$ $color(white)(aaaa)$ $5/3$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $2w+3$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $3w-5$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $f(w)$ $color(white)(aaaaaaa)$ $+$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

Therefore,

$f(w)<0$ when $w in]-3/2, 5/3 [$

How do you solve 3(2w^2-5)<w3(2w^2-5)<w using a sign chart?