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Question Number 110596 by Aina Samuel Temidayo last updated on 29/Aug/20
Three real numbers a,b,c satisfying ab+c=10,bc+a=11,ca+b=14. Find (a−b)(b−c)(c−a)(a−1)(b−1)(c−1)
$$\mathrm{Three}\:\mathrm{real}\:\mathrm{numbers}\:\mathrm{a},\mathrm{b},\mathrm{c}\:\mathrm{satisfying} \\ $$$$\mathrm{ab}+\mathrm{c}=\mathrm{10},\mathrm{bc}+\mathrm{a}=\mathrm{11},\mathrm{ca}+\mathrm{b}=\mathrm{14}.\:\mathrm{Find} \\ $$$$\left(\mathrm{a}−\mathrm{b}\right)\left(\mathrm{b}−\mathrm{c}\right)\left(\mathrm{c}−\mathrm{a}\right)\left(\mathrm{a}−\mathrm{1}\right)\left(\mathrm{b}−\mathrm{1}\right)\left(\mathrm{c}−\mathrm{1}\right) \\ $$
Commented by Her_Majesty last updated on 29/Aug/20
one solution is a=3 b=2 c=4 ⇒ answer is −12
$${one}\:{solution}\:{is} \\ $$$${a}=\mathrm{3}\:{b}=\mathrm{2}\:{c}=\mathrm{4}\:\Rightarrow\:{answer}\:{is}\:−\mathrm{12} \\ $$
Commented by Aina Samuel Temidayo last updated on 29/Aug/20
Thanks but I don′t only need answers. I also need well detailed solutions.
$$\mathrm{Thanks}\:\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{only}\:\mathrm{need}\:\mathrm{answers}. \\ $$$$\mathrm{I}\:\mathrm{also}\:\mathrm{need}\:\mathrm{well}\:\mathrm{detailed}\:\mathrm{solutions}. \\ $$
Answered by Her_Majesty last updated on 30/Aug/20
ab+c=10 bc+a=11 ca+b=14 ===== c=10−ab=((11−a)/b)=((14−b)/a) 10−ab=((14−b)/a) ⇒ b=((2(5a−7))/(a^2 −1)) c=((2(7a−5))/(a^2 −1))=−(((a−11)(a^2 −1))/(2(5a−7))) ((2(7a−5))/(a^2 −1))=−(((a−11)(a^2 −1))/(2(5a−7))) a^5 −11a^4 −2a^3 +162a^2 −295a+129=0 trying factors of 129 which are ±1, ±3, ±43, ±129 ⇒ a_1 =3 ⇒ b_1 =2, c_1 =4 now we must solve this: a^4 −8a^3 −26a^2 +84a−43=0 sadly there′s no other exact solution a_2 ≈−4.07488 a_3 ≈.685287 a_4 ≈1.56780 a_5 ≈9.82180 anyway all 5 triplets (a/b/c) give −12
$${ab}+{c}=\mathrm{10} \\ $$$${bc}+{a}=\mathrm{11} \\ $$$${ca}+{b}=\mathrm{14} \\ $$$$===== \\ $$$${c}=\mathrm{10}−{ab}=\frac{\mathrm{11}−{a}}{{b}}=\frac{\mathrm{14}−{b}}{{a}} \\ $$$$\mathrm{10}−{ab}=\frac{\mathrm{14}−{b}}{{a}}\:\Rightarrow\:{b}=\frac{\mathrm{2}\left(\mathrm{5}{a}−\mathrm{7}\right)}{{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$${c}=\frac{\mathrm{2}\left(\mathrm{7}{a}−\mathrm{5}\right)}{{a}^{\mathrm{2}} −\mathrm{1}}=−\frac{\left({a}−\mathrm{11}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{5}{a}−\mathrm{7}\right)} \\ $$$$\frac{\mathrm{2}\left(\mathrm{7}{a}−\mathrm{5}\right)}{{a}^{\mathrm{2}} −\mathrm{1}}=−\frac{\left({a}−\mathrm{11}\right)\left({a}^{\mathrm{2}} −\mathrm{1}\right)}{\mathrm{2}\left(\mathrm{5}{a}−\mathrm{7}\right)} \\ $$$${a}^{\mathrm{5}} −\mathrm{11}{a}^{\mathrm{4}} −\mathrm{2}{a}^{\mathrm{3}} +\mathrm{162}{a}^{\mathrm{2}} −\mathrm{295}{a}+\mathrm{129}=\mathrm{0} \\ $$$${trying}\:{factors}\:{of}\:\mathrm{129}\:{which}\:{are}\:\pm\mathrm{1},\:\pm\mathrm{3},\:\pm\mathrm{43},\:\pm\mathrm{129} \\ $$$$\Rightarrow\:{a}_{\mathrm{1}} =\mathrm{3} \\ $$$$\Rightarrow\:{b}_{\mathrm{1}} =\mathrm{2},\:{c}_{\mathrm{1}} =\mathrm{4} \\ $$$${now}\:{we}\:{must}\:{solve}\:{this}: \\ $$$${a}^{\mathrm{4}} −\mathrm{8}{a}^{\mathrm{3}} −\mathrm{26}{a}^{\mathrm{2}} +\mathrm{84}{a}−\mathrm{43}=\mathrm{0} \\ $$$${sadly}\:{there}'{s}\:{no}\:{other}\:{exact}\:{solution} \\ $$$${a}_{\mathrm{2}} \approx−\mathrm{4}.\mathrm{07488} \\ $$$${a}_{\mathrm{3}} \approx.\mathrm{685287} \\ $$$${a}_{\mathrm{4}} \approx\mathrm{1}.\mathrm{56780} \\ $$$${a}_{\mathrm{5}} \approx\mathrm{9}.\mathrm{82180} \\ $$$${anyway}\:{all}\:\mathrm{5}\:{triplets}\:\left({a}/{b}/{c}\right)\:{give}\:−\mathrm{12} \\ $$

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