I don't know where to stop...!!!

**Alienis Back** · November 28th 2008, 03:27 AM

Please, see the attached document, for I was unable to write it using Latex.

**mr fantastic** · November 28th 2008, 04:04 AM

Originally Posted by Alienis Back

Please, see the attached document, for I was unable to write it using Latex.

The question is:

How to solve $9x^2 + 6x - 8 = 0$ .

The approach you're trying to use is completing the square. You've done well to get to here:

$9\left[ \left(x + \frac{1}{3}\right)^2 - 1 \right] = 0$ .

Unfortunately you then proceed to undo all this, effectively going in a circle and arriving back to $9x^2 + 6x - 8 = 0$ .

Here is what you must do. Divide both sides of $9\left[ \left(x + \frac{1}{3}\right)^2 - 1 \right] = 0$ by 9:

$\left(x + \frac{1}{3}\right)^2 - 1 = 0$

$\Rightarrow \left(x + \frac{1}{3}\right)^2 - 1^2 = 0$

which is clearly a difference of two squares.

From your previous posts I know that you (should) know how to factorise this and hence get the solution.

**great_math** · November 28th 2008, 04:24 AM

How to solve

.

A better approach...

$9x^2+6x-8=0$
Multiply a 4 to LHS and RHS

It becomes $36x^2+24x-32=0$

$\implies(6x)^2+2\cdot 6x\cdot 2+4-36=0$

$\implies(6x+2)^2-6^2=0$

$\implies (6x-4)(6x+8)=0$

$\implies x=\left\{\frac 2{3},\frac {-4}{3}\right\}$

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