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

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Answer 1 · 2015-07-19T12:04:09

Complete the square by adding #4# to both sides of the equation to find:

#(x+2)^2 = x^2+4x+4 = 5+4 = 9 = 3^2#

Hence #x = 1# or #x=-5#

Given #ax^2+bx#, note that

#a(x+b/(2a))^2 = ax^2+bx+b^2/(4a)#

In our case #a = 1# and #b = 4#, so #b/(2a) = 4/2 = 2# and

#(x+2)^2 = x^2+4x+4#

So add #4# to both sides of our original equation to get:

#x^2+4x+4 = 9#

That is:

#(x+2)^2 = 3^2#

So

#x+2 = +-sqrt(3^2) = +-3#

Subtract #2# from both sides to get:

#x = -2 +-3#