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 $a=2b=c$ ?

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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 $a=2b=c$ ?

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Answer 1 · 2017-12-05T04:05:11

$a=2b=3c$ , 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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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

Explanation:

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The real numbers a, ba, b and cc satisfy the equation: 3a^2 + 4b^2 + 18c^2 - 4ab - 12ac = 03a^2 + 4b^2 + 18c^2 - 4ab - 12ac = 0. By forming perfect squares, how do you prove that a=2b=ca=2b=c?

1 Answer

Explanation:

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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 $a=2b=c$ ?