# Solving Inequalities

• Jan 13th 2008, 04:53 PM
Macleef
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?
• Jan 13th 2008, 08:07 PM
janvdl
Quote:

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
• Jan 13th 2008, 08:21 PM
mr fantastic
Quote:

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.