Question

8x=4x^2-1 solve by completing the square

Answer 1

Get the x terms by themselves on one side and a constant on the other side of the equal sign...

4x^2-8x=1  make the leading coefficient equal to one...

x^2-2x=1/4  now halve the linear coefficient, -2 in this case, square it, and add that value to both sides of the equation...-2/2=-1, -1^2=1 so

x^2-2x+1=1+1/4

x^2-2x+1=5/4  now the left side is a perfect square...

(x-1)^2=5/4  take the square root of both sides...

x-1=±√(5/4)  add 1 to both sides

x=1±√(5/4)

Answer 2

Isolate the 1: 4x^2-8x=1
Since "a" is a perfect square use the formula b^2/4a: (-8)^2/4(4)= 4
Add 4 into both sides of the equation:
4x^2-8x+4=1+4
Ur new equation should be:
4x^2 - 8x + 4 = 5
Factor: (2x-2)^2=5
Solve for x by getting the square root of 5 to get rid of the square on the (2x-2)^2 side and you would get:
2x-2 = +-2.2
solve for x: x= 2.1 x= -0.1