a,b, and c are positive real numbers such that a^2 +ab+(b^2 /3) = 25 , (b^2 /3) + c^2 =9 and c^2 +ca+a^2 = 16 find the value of ab+2bc+3ac


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Question Number 171388 by infinityaction last updated on 15/Jun/22

$a, b, a n d c a r e p o s i t i v e r e a l n u m b e r s$ $s u c h t h a t$ $a^{2} + a b + \frac{b^{2}}{3} = 25, \frac{b^{2}}{3} + c^{2} = 9$ $a n d c^{2} + c a + a^{2} = 16$ $f i n d t h e v a l u e o f a b + 2 b c + 3 a c$
	Answered by MJS_new last updated on 15/Jun/22

	$let b = p a \land c = q a$ $(\frac{p^{2}}{3} + p + 1) a^{2} = 25$ $(\frac{p^{2}}{3} + q^{2}) a^{2} = 9$ $(q^{2} + q + 1) a^{2} = 16 ()$ $\Rightarrow$ $\frac{p^{2} + 3 p + 3}{75} = \frac{p^{2} + 3 q^{2}}{27} = \frac{q^{2} + q + 1}{16}$ $\Rightarrow$ $p = q (2 q + 1)$ $q^{4} + q^{3} + \frac{37}{64} q^{2} - \frac{27}{64} q - \frac{27}{64} = 0$ $(q^{2} - \frac{27}{64}) (q^{2} + q + 1) = 0$ $() \Rightarrow q^{2} + q + 1 \neq 0$ $q = \pm \frac{3 \sqrt{3}}{8} \Rightarrow p = \frac{27}{32} \pm \frac{3 \sqrt{3}}{8}$ $\Rightarrow insert and solve for a$ $. . .$ $answer is \pm 24 \sqrt{3}$
	Commented by infinityaction last updated on 16/Jun/22

	$t h a n k y o u s i r$
	Commented by MJS_new last updated on 16/Jun/22

	$sorry while typing I forgot a, b, c are positive$ $numbers . \Rightarrow the only solution is + 24 \sqrt{3}$
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