How do you solve #x^2+15x+56=0#?

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Answer 1 · 2016-03-27T14:30:02

#x = -7# or #x = -8#

Find a pair of factors of #56# with sum #15#.

The pair #7, 8# works in that #7*8 = 56# and #7+8=15#

Hence:

#0 = x^2+15x+56 = (x+7)(x+8)#

So:

#x = -7# or #x = -8#

#color(white)()#
Alternative method

I will use the difference of squares identity:

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

with #a=(2x+15)# and #b=1# as follows:

First multiply the whole equation by #4# (to cut down on the use of fractions) to get:

#0 = 4x^2+60x+224#

#=(2x+15)^2-225+224#

#=(2x+15)^2-1^2#

#=((2x+15)-1)((2x+15)+1)#

#=(2x+14)(2x+16)#

#=(2(x+7))(2(x+8))#

#=4(x+7)(x+8)#

Hence #x = -7# or #x = -8#