completing the square: 4x^2 - 12x + 7 = 0

[haliebre]

When completing the square what number is added to each side of the equation w
4x^2-12x+7=0 I came up with 3 is that right? I first divided the equation by 4 and then I took had x^2 -3x + 7/4=0 then I took 1/2 of 3 to get 1.5 and squared it to get 3?

 

[pka]

Please note how we factor the square to a monic inside the parentheses:
\(\displaystyle 4\left( {x^2 - 3x + \quad \quad } \right) = - 7\).

Complete the square inside the parentheses:
\(\displaystyle 4\left( {x^2 - 3x + \frac{9}{4}} \right) = - 7 + 9\), noting we have added just 9 to the expression.

Thus we have \(\displaystyle \begin{array}{l}
4\left( {x - \frac{3}{2}} \right)^2 = 2 \\
\left( {x - \frac{3}{2}} \right)^2 = \frac{1}{2} \\
\end{array}\)

 

[soroban]

Re: completing the square

Hello, haliebre!

Close, but two errors . . .

When completing the square what number is added to each side of the equation:
\(\displaystyle \;\;\;\;4x^2\,-\,12x\,+\,7\:=\:0\;\) [1]

I came up with 3 is that right? \(\displaystyle \;\) . . . no
I first divided the equation by 4: \(\displaystyle \,x^2\,-\,3x\,+\,\frac{7}{4}\:=\:0\;\) [2]
then I took \(\displaystyle \frac{1}{2}\) of \(\displaystyle 3\) to get \(\displaystyle 1.5\;\)
and squared it to get 3? \(\displaystyle \;\) . . . oops!

You doubled it . . . \(\displaystyle \,1.5^2\,=\,2.25\)

However, this is not the answer to the problem!
This is the number added to both sides of your equation, [2].

pka suggested the correct procedure.

 

[haliebre]

Now I have a real dilemma. My choices for answers to this equation are:

a. 9/4
b. 36
c. 9/2
d. 3

 

[Gene]

No dilemma. PKA added 9/4