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

• Nov 28th 2008, 03:27 AM
Alienis Back
I don't know where to stop...!!!
Please, see the attached document, for I was unable to write it using Latex.
• Nov 28th 2008, 04:04 AM
mr fantastic
Quote:

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 $\displaystyle 9x^2 + 6x - 8 = 0$.

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

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

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

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

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

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

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.
• Nov 28th 2008, 04:24 AM
great_math
How to solve http://www.mathhelpforum.com/math-he...1ed212fd-1.gif.

A better approach...

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

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

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

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

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

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