complete the square

**decoy808** · November 29th 2009, 06:08 AM

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

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

$3x^2+2x=8$

$<br /> x^2+\frac{2x}{3}=\frac{8}{3}<br />$

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

$<br /> x^2+\frac{2x}{3}+\frac{1}{4}=\frac{8}{3}+\frac{1}{ 4}<br />$

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

thanks

**Raoh** · November 29th 2009, 06:17 AM

hello

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

.

**decoy808** · November 29th 2009, 06:28 AM

hello

okey so my step with 1/4 is wrong?

**Raoh** · November 29th 2009, 06:34 AM

Originally Posted by decoy808

hello

okey so my step with 1/4 is wrong?

looks that way

.

**Raoh** · November 29th 2009, 06:56 AM

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 $\frac{2}{3}x$ must be divided by 2 not $\frac{1}{2}$ .

**decoy808** · November 29th 2009, 07:33 AM

haha do not trust computers

my answer is x=2 or -4/3

am i correct? many thanks

**Raoh** · November 29th 2009, 08:03 AM

Almost

, $x=-2$ or $x=\frac{4}{3}$

**Raoh** · November 29th 2009, 08:08 AM

anyway, if the problem was "solve the equation",it would have been better if you factorized

.
$3x^2+2x-5=(x+2)(3x-4)$

**decoy808** · November 29th 2009, 08:10 AM

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

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