Given:
The equation is:
[tex]x^2-12x+32=4[/tex]
To find:
The result of completing the square for the given equation.
Solution:
We have,
[tex]x^2-12x+32=4[/tex]
It can be written as:
[tex]x^2-12x=4-32[/tex]
[tex]x^2-12x=-28[/tex]
Now, we need to add half of square of coefficient of x, to complete the square.
Adding [tex](\dfrac{-12}{2})^2[/tex] on both sides, we get
[tex]x^2-12x+(\dfrac{-12}{2})^2=-28+(\dfrac{-12}{2})^2[/tex]
[tex]x^2-2(x)(6)+(-6)^2=-28+(-6)^2[/tex]
[tex]x^2-2(x)(6)+(6)^2=-28+36[/tex] [tex][\because (-a)^2=(a)^2][/tex]
Using the formula [tex](a-b)^2=a^2-2ab+b^2[/tex], we get
[tex](x-6)^2=8[/tex]
Therefore, the required equation after completing the square is [tex](x-6)^2=8[/tex].
