1. ## complete the square

i have a worked problem with solution, but need to check.

$\displaystyle 3x^2+2x-8=0$

$\displaystyle 3x^2+2x=8$

$\displaystyle x^2+\frac{2x}{3}=\frac{8}{3}$

now i must take half coefficient of x and square,right?
the next step i have been given is

$\displaystyle x^2+\frac{2x}{3}+\frac{1}{4}=\frac{8}{3}+\frac{1}{ 4}$

but i cannot see where the 1/4 is coming from?
as 2x/3 divided by 1/2 = 1/3?

thanks

2. hello
$\displaystyle 3x^2+2x-8=3\left ( x^2+\frac{2}{3}x-\frac{8}{3} \right )$=$\displaystyle 3\left (\left ( x+\frac{2}{6} \right )^2-\frac{4}{36}-\frac{8}{3} \right )$$\displaystyle =3\left ( (x+\frac{2}{6})^2-\frac{25}{9} \right )$.
u do some simplifications.

3. hello okey so my step with 1/4 is wrong?

4. Originally Posted by decoy808
hello okey so my step with 1/4 is wrong?
looks that way.

5. Originally Posted by decoy808
hello okey so my step with 1/4 is wrong?
i don't see where 1/4 is coming from, but $\displaystyle \frac{2}{3}x$ must be divided by 2 not $\displaystyle \frac{1}{2}$.

6. haha do not trust computers
my answer is x=2 or -4/3

am i correct? many thanks

7. Almost , $\displaystyle x=-2$ or $\displaystyle x=\frac{4}{3}$

8. anyway, if the problem was "solve the equation",it would have been better if you factorized .
$\displaystyle 3x^2+2x-5=(x+2)(3x-4)$

9. i take it back....garbage in, garbage out