Question

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

Answer 1

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

Explanation:

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]}