Solve by Completing the Square?

Question

Solve this problem by completing the square:

x^2-4x=7

if you could show work too that would be great! thanks!

Answer 1

x^2-4x=7

x^2-4x-7=0

delta=b^2-4ac

         =(4^2)-4(1*-7)

        =16-(-28)

        =16+28

        =44

sq root of 44=

i dnt have calculator near me

x'=-b-sqrootdelta/2(1)

x''=-b+sq root delta/2(1)

you need to find square root delta and replace it wherever you see sqroot delta

Answer 2

Completing the square means writing an equation in vertex form.  From this form you can identify the vertex and the range.  I'm not sure whether you want to complete the square or just solve the equation.  You didn't put the variable "y" in the problem.  I'll answer it as if it were

x^2 - 4x = y + 7

Grab the 4, take half of it, then square that result.

4/2 = (2) ^ 2 = 4

x^2 - 4x + 4 = y + 7 + 4 (you are adding 4 to both sides, which makes the left side a perfect square trinomial, hence the name completing the square)

(x - 2) ^ 2 = y + 11

If you are trying to SOLVE x^2 - 4x = 7 then I would use quadratic formula instead.

Good luck to you !!

Answer 3

x^2 - 4x = 7

x^2 - 4x + 4 = 7 + 4     (add 4 for both sides)

(x - 2)^2    = [sqrt(11)]^2

or (x - 2 ) = +/- sqrt(11)

****  x - 2 = +sqrt(11) ==> x = 2 +sqrt(11

OR

****  x - 2 = -sqrt(11)  ==> x = 2 - aqrt(11)