technique of completing the square to transform the quadratic equation

May 2010
43
0
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?
 

Pim

Dec 2008
91
39
The Netherlands
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 2010
43
0
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.
 

Grandad

MHF Hall of Honor
Dec 2008
2,570
1,416
South Coast of England
Hello everyone

I'm sure Pim meant to say:

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

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

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

Grandad
 
May 2010
43
0
Hello everyone

I'm sure Pim meant to say:
\(\displaystyle 4x^2+16x+12=0\)
\(\displaystyle \Rightarrow x^2+4x+3=0\)

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

\(\displaystyle \Rightarrow (x+2)^2=1\)
Grandad
Thanks I was checking up on it to see what I got and it was right. Thanks again.