Answer:
\(x=6+\sqrt{31}\)
\(x=6-\sqrt{31}\)
Step-by-step explanation:
Solve \(x^2-12x+5 = 0\) using the completing-the-square method.
Subtract 5 from both sides
\(x^2-12x=-5\)
In completing the square method , we take coefficient of middle term and then divide by 2 and square it
12 divide by 2 is 6
6^2 = 36
Add 36 on both sides
\(x^2-12x+36=-5+36\)
\(x^2-12x+36=31\)
Now we factor the left hand side
\((x-6)(x-6)=31\)
\((x-6)^2=31\)
Take square root on both sides
\(x-6=+-\sqrt{31}\)
Now add 6 on both sides
\(x=6+-\sqrt{31}\)
So there are two values for x
\(x=6+\sqrt{31}\)
\(x=6-\sqrt{31}\)
