How do you solve $2x^2+3x>0$ using a sign chart?

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Answer 1 · 2017-02-10T17:10:53

The answer is $x in ]-oo, -3/2[ uu ]0, +oo[$

Let $f(x)=2x^2+3x=x(2x+3)$

We now, build the sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaaaa)$ $-oo$ $color(white)(aaaa)$ $-3/2$ $color(white)(aaaa)$ $0$ $color(white)(aaaa)$ $+oo$

$color(white)(aaaa)$ $2x+3$ $color(white)(aaaaaaa)$ $-$ $color(white)(aaaa)$ $+$ $color(white)(aaaa)$ $+$

$color(white)(aaaa)$ $x$ $color(white)(aaaaaaaaaaa)$ $-$ $color(white)(aaaa)$ $-$ $color(white)(aaaa)$ $+$

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

Therefore,

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

How do you solve 2x^2+3x>02x^2+3x>0 using a sign chart?