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-

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

\dots

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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