Solving Inequalities

**Macleef** · January 13th 2008, 04:53 PM

How would you solve the following inequality algebraically?

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

Is it. . .

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

And use interval charts?

**janvdl** · January 13th 2008, 08:07 PM

Originally Posted by Macleef

How would you solve the following inequality algebraically?

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

Is it. . .

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

And use interval charts?

Multiply by 2 throughout the equation

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

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

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

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

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

**mr fantastic** · January 13th 2008, 08:21 PM

Originally Posted by janvdl

Multiply by 2 throughout the equation Mr F explains: To make life easier.

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

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

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

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

The roots of this equation still aren't exactly nice ones to work with .....

Alterations and paint work by Mr F.

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