# technique of completing the square to transform the quadratic equation

• May 18th 2010, 07:35 AM
technique of completing the square to transform the quadratic equation
Hey guys, this is a nice forum Im glad there is one out there that can help people with them skills.

Use the technique of completing the square to transform the quadratic equation below into the form (x + c)2 = a.

4x2 + 16x + 12 = 0

I seem to be having a bit of trouble figuiring this out...can someone help?
• May 18th 2010, 07:43 AM
Pim
firstly, you'll want to get rid of the 4

$\displaystyle 4x^2+16x -12 = 0$
$\displaystyle x^2+4x-3 = 0$

Then, ask yourself, if you work out the brackets of $\displaystyle (x+c)^2$, what makes sure you get +4x ? In this case that is $\displaystyle c=2 \frac{4}{2}$, because $\displaystyle (x+2)^2 = x^2 + 4x + 4$
So, what you have now is:
$\displaystyle (x+2)^2 = x^2 + 4x + 4$
$\displaystyle (x+2)^2 - 4 = x^2 + 4x$
$\displaystyle (x+2)^2 -7 = x^2 + 4x - 3$
Therefore, $\displaystyle (x+2)^2 = 7$ is equivalent to the first equation.

Does that make it clear?
• May 18th 2010, 07:47 AM
Quote:

Originally Posted by Pim
firstly, you'll want to get rid of the 4

$\displaystyle 4x^2+16x -12 = 0$
$\displaystyle x^2+4x-3 = 0$

Then, ask yourself, if you work out the brackets of $\displaystyle (x+c)^2$, what makes sure you get +4x ? In this case that is $\displaystyle c=2 \frac{4}{2}$, because $\displaystyle (x+2)^2 = x^2 + 4x + 4$
So, what you have now is:
$\displaystyle (x+2)^2 = x^2 + 4x + 4$
$\displaystyle (x+2)^2 - 4 = x^2 + 4x$
$\displaystyle (x+2)^2 -7 = x^2 + 4x - 3$
Therefore, $\displaystyle (x+2)^2 = 7$ is equivalent to the first equation.

Does that make it clear?

Thanks. Well sorta but the answer doesn't fit the from whats above.
• May 18th 2010, 07:50 AM
Hello everyone

I'm sure Pim meant to say:

$\displaystyle 4x^2+16x+12=0$
$\displaystyle \Rightarrow x^2+4x+3=0$

$\displaystyle \Rightarrow (x+2)^2-1=0$

$\displaystyle \Rightarrow (x+2)^2=1$

• May 18th 2010, 08:22 AM
Quote:

$\displaystyle 4x^2+16x+12=0$
$\displaystyle \Rightarrow x^2+4x+3=0$
$\displaystyle \Rightarrow (x+2)^2-1=0$
$\displaystyle \Rightarrow (x+2)^2=1$