How do you solve using the completing the square method #3x^2-5x-6=0#?

1 Answer
George C.
May 10, 2016

#x = 5/6+-sqrt(97)/6#

Explanation:

Use the difference of squares identity:

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

with #a=(6x-5)# and #b=sqrt(97)#.

To avoid too much arithmetic with fractions, pre-multiply the equation by #3*2^2 = 12# first:

#0 = 12(3x^2-5x-6)#

#=36x^2-60x-72#

#=(6x-5)^2-5^2-72#

#=(6x-5)^2-25-72#

#=(6x-5)^2-97#

#=(6x-5)^2-(sqrt(97))^2#

#=((6x-5)-sqrt(97))((6x-5)+sqrt(97))#

#=(6x-5-sqrt(97))(6x-5+sqrt(97))#

#=36(x-5/6-sqrt(97)/6)(x-5/6+sqrt(97)/6)#

So #x = 5/6+-sqrt(97)/6#

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