Question

How do you solve `2x^2+4x<=3` using a sign chart?

Answer 1

The answer is `x in [-2.58, 0.58]`

Explanation:

Let`s rewrite the equation

`2x^2+4x-3<=0`

We need the roots of the equation

`2x^2+4x-3=0`

We calculate the discriminant

`Delta=b^2-4ac=16+4*2*3=40`

As `Delta>0`, there are 2 real roots

Therefore,

`x_1=(-4+sqrt40)/4=0.58`

`x_2=(-4-sqrt40)/4=-2.58`

Let `f(x)=(x+2.58)(x-0.58)<=0`

Now, we can do our sign chart

`color(white)(aaaa)` `x` `color(white)(aaaa)` `-oo` `color(white)(aaaa)` `-2.58` `color(white)(aaaaa)` `0.58` `color(white)(aaaa)` `+oo`

`color(white)(aaaa)` `x+2.58` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `+` `color(white)(aaaaaa)` `+`

`color(white)(aaaa)` `x-0.58` `color(white)(aaaa)` `-` `color(white)(aaaaa)` `-` `color(white)(aaaaaa)` `+`

`color(white)(aaaa)` `f(x)` `color(white)(aaaaaaaa)` `+` `color(white)(aaaaa)` `-` `color(white)(aaaaaa)` `+`

Therefore,

`f(x)<=0`, when `x in [-2.58, 0.58]`