How do you solve $2x^2+5x-12>0$ using a sign chart?

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Answer 1 · 2016-12-01T06:39:17

The answer is $x in] -oo,-4 [ uu ] 3/2, oo[$

Let's start by solving the quadratic equation

$ax^2+bx+c=0$

Our equation is $f(x)=2x^2+5x-12$

Let's calculate the discriminant

$Delta=b^2-4ac=25-4*2(-12)=121$

$x=(-b+-sqrtDelta)/(2a)$

$x=(-5+-sqrt121)/4$

$x=(-5+-11)/4$

$x_1=-16/4=-4$

$x_2=6/4=3/2$

Let's do our sign chart

$color(white)(aaaa)$ $x$ $color(white)(aaaa)$ $-oo$ $color(white)(aaaa)$ $-4$ $color(white)(aaaa)$ $3/2$ $color(white)(aaaa)$ $+oo$

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

$color(white)(aaaa)$ $x-3/2$ $color(white)(aaaa)$ $-$ $color(white)(aaaaa)$ $-$ $color(white)(aaaa)$ $+$

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

So, $f(x)>0$ , when $x in] -oo,-4 [ uu ] 3/2, oo[$

How do you solve 2x^2+5x-12>02x^2+5x-12>0 using a sign chart?