Question

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

Answer 1

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

Explanation:

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[`