How do you solve #9x^2-12x-14=0# by completing the square?

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Answer 1 · 2016-08-13T04:06:40

#x=2/3-sqrt2# or #x=2/3+sqrt2#

In #9x^2-12x-14=0#, while #9x^2=(3x)^2#, to complete the square, recall the identity #(x+-a)^2=x^2+-2ax+a^2#.

As #-12x=-2×(3x)×2#, we need to add #2^2# to make it complete square.

Hence, #9x^2-12x-14=0# can be written as #((3x)^2-2×(3x)×2+2^2)-4-14=0# or

#(3x-2)^2-18=0#, which is equivalent to

#(3x-2)^2-(sqrt18)^2=0#

and using identity #(a-b)^2=(a+b)(a-b)# we can write the equation as

#(3x-2+sqrt18)(3x-2-sqrt18)=0#

i.e. either #3x-2+sqrt18=0# or #3x-2-sqrt18=0#.

Now as #sqrt18=3sqrt2#

either #x=2/3-sqrt2# or #x=2/3+sqrt2#