How do you solve the inequality $2x^2+25x+63<=0$ ?

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Answer 1 · 2017-02-01T14:02:53

The answer is $x in [-9, -7/2]$

Let's factorise the LHS

$2x^2+25x+63<=0$

$(2x+7)(x+9)<=0$

Let $f(x)=(2x+7)(x+9)$

We can build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-9$ $color(white)(aaaa)$ $-7/2$ $color(white)(aaaa)$ $-oo$

$color(white)(aaaa)$ $x+9$ $color(white)(aaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $2x+7$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore,

$f(x)<=0$ , when $x in [-9, -7/2]$

How do you solve the inequality 2x^2+25x+63<=02x^2+25x+63<=0?