How do you solve #x^2+4x=5# by completing the square?

2 Answers
Jul 5, 2015

Answer:

#color(green)(x=1, x=-5#

Explanation:

#x^2 + 4x =5#

To write the Left Hand Side as a Perfect Square, we add 4 to both sides
#x^2 + 4x + 4= 5 + 4#

#x^2 + 2*2*x + 2^2 = 9#

Using the Identity #color(blue)((a+b)^2 = a^2 + 2ab + b^2#, we get
#(x+2)^2 = 9#

#x + 2 = sqrt9# or #x +2 = -sqrt9#

#color(green)(x = 3 -2 =1# or
# color(green)(x = -3-2=-5#

The solutions are
#color(green)(x=1, x=-5#

Jul 5, 2015

Answer:

Adjust the left side so that it becomes a perfect square.

Explanation:

Completing the square only works when you don't have a coefficient in front of #x^2# (i.e., a coefficient of 1). If you did have a coefficient other than one, you would have needed to divide every part of the expression by that value to make it one.

The next step is to look at the coefficient of #x#, in this case 4. You have to cut it in half, then square it, and add it to both sides of your original equation:

#(1/2 * 4)^2= 4#

Remember: whatever you add to one side, you need to add to the other.

#x^2 + 4x + 4 = 5 + 4#
#(x+2)^2=9# completing the square makes a perfect square
#x+2 = +-sqrt(9)# take #+-# the square root of both sides
#x+2=+-3# evaluate the square root
#x=-2+-3# subtract two from both sides

So the solution set is is #[-5, 1]#