how do you solve x^2-12x+5=0 by completing the square?

Answer 1

To complete the square follow these steps:

First, get the unit coefficient (5) out of the way, by subtracting 5 from both sides:
x^2 - 12x = -5

Now take the x term coefficient (-12), halve it (-6) and square it (36). This is the number you need to add to both sides:
x^2 - 12x + 36 = 31

Now the left hand side can be written as a perfect square:
(x - 6)^2 = 31

Now you can take the square root of both sides and solve for x:

x-6 = sqrt(31)
or
x-6 = -sqrt(31)

So x = 6 +/- sqrt(31)

Answer 2

get rid of 5 on both sides

so that you get this (x^2)-12x = -5

divide the x variable which is 12 by 2 so you get six
2 (square) the 6 so you get 36

add 36 on both sides
so you get ,

(x^2)-12x+36 = 31
then you simplify and finish the problem which comes out too.....

(x-6)^2 = 31

Answer 3

1x^2-12x+5 = 0
b"2-4ac = 12"2-4*1*5 = 124
x = (-(-12)+- sqrt( 12^2 -4*1*5 ))/(2*1)
x = (-(-12)+- sqrt( 124 ) )/(2*1)
6+- sqrt( 31)

its been a few years i had to take a refresher look at the web page

Answer 4

subtract 5 on both sides
looks like (x^2)-12x = -5

divide the x variable (12) by two so you get 6
square the 6 so you get 36

add 36 on both sides
looks like:
(x^2)-12x+36 = 31
simplify:
(x-6)^2 = 31

Answer 5

x^2 - 12x + 5 = 0
x^2 - 12x = -5
x^2 - 12x + 36 = 31
(x - 6)^2 = 31
x - 6 = sqrt(31)
x = 6 ± sqrt(31)

Answer 6

Add: (b/2)^2 to both sides: (-12/2)^2 = (-6)^2 = 36