How do you solve $2u^2+5u>=12$ using a sign chart?

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How do you solve $2u^2+5u>=12$ using a sign chart?

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Answer 1 · 2017-11-28T10:44:49

The solution is $u in (-oo,-4] uu [3/2, +oo)$

Explanation:

Lets 's factorise the function

$f(u)=2u^2+5u-12=(2u-3)(u+4)$

The equation we have to solve is

$f(u)=(2u-3)(u+4)>=0$

Let's build the sign chart

$color(white)(aaaa)$ $u$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaaaaaaaa)$ $3/2$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $u+4$ $color(white)(aaaaaa)$ $-$ $color(white)(aa)$ $0$ $color(white)(aaaa)$ $+$ $color(white)(aaaaaaa)$ $+$

$color(white)(aaaa)$ $2u-3$ $color(white)(aaaaa)$ $-$ $color(white)(aa)$ #color(white)(aaaaa)-# $color(white)(aaaa)$ $0$ $color(white)(aaa)$ $+$

$color(white)(aaaa)$ $f(u)$ $color(white)(aaaaaaa)$ $+$ $color(white)(aa)$ $0$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $0$ $color(white)(aaa)$ $+$

Therefore,

$f(u)>=0$ when $u in (-oo,-4] uu [3/2, +oo)$

graph{2x^2+5x-12 [-29.27, 28.45, -13.66, 15.22]}

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