How do you solve using completing the square method #x^2+5x-6=0#?

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Answer 1 · 2016-03-14T20:50:58

See explanation...

I will use the difference of squares identity:

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

with #a=(2x+5)# and #b=7# later...

If I see an odd middle coefficient and am asked to "complete the square", I tend to groan slightly inwardly at the prospect of messy fractions.

Happily we can avoid some of the mess by multiplying our equation through by #2^2 = 4# first:

#0 = 4(x^2+5x-6)#

#=4x^2+20x-24#

#=(2x+5)^2-25-24#

#=(2x+5)^2-49#

#=(2x+5)^2-7^2#

#=((2x+5)-7)((2x+5)+7)#

#=(2x-2)(2x+12)#

#=4(x-1)(x+6)#

So the roots are: #x=1# and #x=-6#