How do you solve $2x^2+5x-12>=0$ ?

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Answer 1 · 2016-11-30T15:05:14

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

Let's find the roots of the equation

$2x^2+5x-12=0$

This is a simultaneous equation, $ax^2+bx+c=0$

We calculate the discriminant,

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

$delta>0$ , so we have 2 real roots

The roots are

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

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

$x_1=-4$ and $x_2=6/4=3/2$

Let $f(x)=2x^2+5x-12$

We do a sign chart

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

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

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

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

We need $f(x)>=0$

$x in ] -oo,-4] uu [3/2, oo[$
graph{2x^2+5x-12 [-12.66, 12.66, -6.34, 6.32]}

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