How do you solve $3x^2-11x-4<=0$ using a sign chart?

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Answer 1 · 2017-01-16T12:43:22

The answer is $x in [-1/3,4]$

Let's factorise the expression

$3x^2-11x-4=(3x+1)(x-4)$

Let $f(x)=3x^2-11x-4$

Now, we can make the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaa)$ $-oo$ $color(white)(aaaa)$ $-1/3$ $color(white)(aaaaa)$ $4$ $color(white)(aaaaa)$ $+oo$

$color(white)(aaaa)$ $3x+1$ $color(white)(aaaaa)$ $-$ $color(white)(aaaaa)$ $+$ $color(white)(aaaa)$ $+$

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

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

Therefore,

$f(x)<=0$ when $x in [-1/3,4]$

How do you solve 3x^2-11x-4<=03x^2-11x-4<=0 using a sign chart?