How do you solve #m^2 + 8m + 15 = 0# by completing the square?

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Answer 1 · 2016-04-02T06:43:34

Complete the square to find #m = -3# or #m = -5#

Note that:

#(m+4)^2 = m^2+2(m)(4) + 4^2 = m^2+8m+16#

So add #1# ti both sides of the equation to get:

#m^2+8m+16 = 1#

which we can write as:

#(m+4)^2 = 1#

Then take the square root of both sides, allowing for both positive and negative square roots to find:

#m+4 = +-sqrt(1) = +-1#

Subtract #4# from both sides to get:

#m = -4+-1#

That is #m = -5# or #m = -3#

#color(white)()#
Alternative method

Use the difference of squares identity:

#a^2-b^2 = (a-b)(a+b)#

with #a = (m+4)# and #b=1# as follows:

#0 = m^2+8m+15#

#= (m+4)^2-16+15#

#= (m+4)^2-1#

#= (m+4)^2 - 1^2#

#= ((m+4) - 1)((m+4) + 1)#

#= (m+3)(m+5)#

So #m = -3# or #m = -5#