# Thread: Solving Inequalities

1. ## Solving Inequalities

How would you solve the following inequality algebraically?

$\displaystyle 0.5(x + 2)^2 (x - 4) > 2x + 3$

Is it. . .

$\displaystyle 0.5(x + 2)^2 (x - 4) - 2x + 3 > 0$

And use interval charts?

2. Originally Posted by Macleef
How would you solve the following inequality algebraically?

$\displaystyle 0.5(x + 2)^2 (x - 4) > 2x + 3$

Is it. . .

$\displaystyle 0.5(x + 2)^2 (x - 4) - 2x + 3 > 0$

And use interval charts?
Multiply by 2 throughout the equation

$\displaystyle (x + 2)^2 (x - 4) > 4x + 6$

$\displaystyle (x^2 + 4x + 4) (x - 4) > 4x + 6$

$\displaystyle x^3 - 4x^2 + 4x^2 - 16x + 4x - 16 -4x - 6 > 0$

$\displaystyle x^3 - 16x - 22 > 0$

x = -2,9 ; -1,66 ; 4,5

3. Originally Posted by janvdl
Multiply by 2 throughout the equation Mr F explains: To make life easier.

$\displaystyle (x + 2)^2 (x - 4) > 4x + 6$ Mr F corrects:

$\displaystyle (x^2 + 4x + 4) (x - 4) > 4x$ + 6

$\displaystyle x^3 - 4x^2 + 4x^2 - 16x + 4x - 16 - 4x$ - 6 > 0

$\displaystyle x^3 - 16x$ - 22 > 0

The roots of this equation still aren't exactly nice ones to work with .....
Alterations and paint work by Mr F.