The real numbers #a, b# and #c# satisfy the equation: #3a^2 + 4b^2 + 18c^2 - 4ab - 12ac = 0#. By forming perfect squares, how do you prove that #a=2b=c#?

1 Answer
iceman
Dec 5, 2017

#a=2b=3c# , See the explanation and the proof below.

Explanation:

#3a^2+4b^2+18c^2-4ab-12ac=0#
Notice that the coefficients are all even except for a^2 i.e: 3, rewrite as follow to group for factoring:
#a^2-4ab+4b^2+2a^2-12ac+18c^2=0#
#(a^2-4ab+4b^2)+2(a^2-6ac+9c^2)=0#
#(a - 2b)^2+2(a-3c)^2=0#
We have a perfect square term plus twice perfect square of another term equal to zero, for this to be true each term of the sum must be equal to zero, then:
#(a - 2b)^2=0# and #2(a-3c)^2=0#
#a-2b=0# and #a-3c=0#
#a=2b# and #a=3c#
thus:
#a=2b=3c#
Hence proved.

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