How do you solve $2v^2+7v<4$ using a sign chart?

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Answer 1 · 2016-12-31T07:27:38

The answer is $v in ] -4,1/2 [$

Let's rewrite the inequality as

$2v^2+7v-4<0$

Let $f(v)=2v^2+7v-4$

The domain of $f(v)$ is $D_f(v)=RR$

Let's factorise the expression

$2v^2+7v-4=(2v-1)(v+4)$

Now we establish the sign chart

$color(white)(aaaa)$ $v$ $color(white)(aaaaa)$ $-oo$ $color(white)(aaaaa)$ $-4$ $color(white)(aaaaa)$ $1/2$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $v+4$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaaa)$ $+$

$color(white)(aaaa)$ $2v-1$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaa)$ $+$

$color(white)(aaaa)$ $f(v)$ $color(white)(aaaaaaaa)$ $+$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaa)$ $+$

Therefore,

$f(v)<0$ when $v in ] -4,1/2 [$

How do you solve 2v^2+7v<42v^2+7v<4 using a sign chart?