How do you solve by completing the square: #c²-36= -4c#?

1 Answer
Apr 4, 2015
  • First, we Transpose the Constant to one side of the equation.
    Transposing #-36# to the other side we get:
    #c^2+4c = 36#

  • Application of #(a+b)^2 = a^2 + 2ab + b^2#
    We look at the Co-efficient of #c#. It's #4#
    We take half of this number (including the sign), giving us #2#
    We square this value to get #(2)^2 = 4#. We add this number to BOTH sides of the Equation.
    #c^2+4c+4 = 36+4#
    #c^2+4c+4 = 40#
    The Left Hand side #c^2+4c+4# is in the form #a^2 + 2ab + b^2#
    where #a# is #c#, and #b# is #2#

  • The equation can be written as
    #(c+2)^2 = 40#

So #(c+2)# can take either #2sqrt10# or #-2sqrt10# as a value. That's because squaring both will give us 40.

#c+2 = 2sqrt10# (or) #c+2 = -2sqrt10#
#c = 2sqrt10-2# (or) #c = -2sqrt10-2#
#c = 2(sqrt10-1)# (or) #c = -2(sqrt10+1)#

  • Solution : #c = 2(sqrt10-1) , -2(sqrt10+1) #

  • Verify your answer by substituting these values in the Original Equation #c^2- 36 = -4c#. You will see that the solution is correct.