How do you solve #x^2+x-72=0#?

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Answer 1 · 2015-12-23T11:32:29

Find a pair of factors of #72# which differ by #1#, hence:

#x^2+x-72 = (x+9)(x-8)#

which has zeros #x=8# and #x=-9#

In general #(x+a)(x-b) = x^2+(a-b)x-ab#

In our case we want #ab=72# and #a-b = 1#

It should not take long to find that #a=9# and #b=8# works.

Answer 2 · 2015-12-23T11:39:03

Complete the square to find zeros #x=8# and #x=-9#

Alternatively, complete the square as follows:

Note that #(x+1/2)^2 = x^2+x+1/4#

We also use the difference of squares identity:

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

with #a = x+1/2# and #b=17/2#

So:

#x^2+x-72#

#= x^2+x+1/4-1/4-72 = (x+1/2)^2-289/4#

#= (x+1/2)^2-(17/2)^2#

#= ((x+1/2)-17/2)((x+1/2)+17/2)#

#= (x-8)(x+9)#

which has zeros #x=8# and #x=-9#