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

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

$${a},{b},\:{and}\:{c}\:{are}\:{positive}\:{real}\:{numbers} \\ $$$${such}\:{that} \\ $$$${a}^{\mathrm{2}} +{ab}+\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\:=\:\mathrm{25}\:\:,\:\:\:\:\frac{{b}^{\mathrm{2}} }{\mathrm{3}}\:+\:{c}^{\mathrm{2}} \:=\mathrm{9} \\ $$$${and}\:{c}^{\mathrm{2}} +{ca}+{a}^{\mathrm{2}} \:=\:\mathrm{16} \\ $$$${find}\:\:{the}\:\:{value}\:{of}\:\:{ab}+\mathrm{2}{bc}+\mathrm{3}{ac} \\ $$

Answered by MJS_new last updated on 15/Jun/22

let b=pa∧c=qa ((p^2 /3)+p+1)a^2 =25 ((p^2 /3)+q^2 )a^2 =9 (q^2 +q+1)a^2 =16 (∗) ⇒ ((p^2 +3p+3)/(75))=((p^2 +3q^2 )/(27))=((q^2 +q+1)/(16)) ⇒ p=q(2q+1) q^4 +q^3 +((37)/(64))q^2 −((27)/(64))q−((27)/(64))=0 (q^2 −((27)/(64)))(q^2 +q+1)=0 (∗) ⇒ q^2 +q+1≠0 q=±((3(√3))/8) ⇒ p=((27)/(32))±((3(√3))/8) ⇒ insert and solve for a ... answer is ±24(√3)

$$\mathrm{let}\:{b}={pa}\wedge{c}={qa} \\ $$$$\left(\frac{{p}^{\mathrm{2}} }{\mathrm{3}}+{p}+\mathrm{1}\right){a}^{\mathrm{2}} =\mathrm{25} \\ $$$$\left(\frac{{p}^{\mathrm{2}} }{\mathrm{3}}+{q}^{\mathrm{2}} \right){a}^{\mathrm{2}} =\mathrm{9} \\ $$$$\left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right){a}^{\mathrm{2}} =\mathrm{16}\:\left(\ast\right) \\ $$$$\Rightarrow \\ $$$$\frac{{p}^{\mathrm{2}} +\mathrm{3}{p}+\mathrm{3}}{\mathrm{75}}=\frac{{p}^{\mathrm{2}} +\mathrm{3}{q}^{\mathrm{2}} }{\mathrm{27}}=\frac{{q}^{\mathrm{2}} +{q}+\mathrm{1}}{\mathrm{16}} \\ $$$$\Rightarrow \\ $$$${p}={q}\left(\mathrm{2}{q}+\mathrm{1}\right) \\ $$$${q}^{\mathrm{4}} +{q}^{\mathrm{3}} +\frac{\mathrm{37}}{\mathrm{64}}{q}^{\mathrm{2}} −\frac{\mathrm{27}}{\mathrm{64}}{q}−\frac{\mathrm{27}}{\mathrm{64}}=\mathrm{0} \\ $$$$\left({q}^{\mathrm{2}} −\frac{\mathrm{27}}{\mathrm{64}}\right)\left({q}^{\mathrm{2}} +{q}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left(\ast\right)\:\Rightarrow\:{q}^{\mathrm{2}} +{q}+\mathrm{1}\neq\mathrm{0} \\ $$$${q}=\pm\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}}\:\Rightarrow\:{p}=\frac{\mathrm{27}}{\mathrm{32}}\pm\frac{\mathrm{3}\sqrt{\mathrm{3}}}{\mathrm{8}} \\ $$$$\Rightarrow\:\mathrm{insert}\:\mathrm{and}\:\mathrm{solve}\:\mathrm{for}\:{a} \\ $$$$... \\ $$$$\mathrm{answer}\:\mathrm{is}\:\pm\mathrm{24}\sqrt{\mathrm{3}} \\ $$

Commented by infinityaction last updated on 16/Jun/22

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by MJS_new last updated on 16/Jun/22

sorry while typing I forgot a, b, c are positive numbers. ⇒ the only solution is +24(√3)

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

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