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Question Number 152678 by mathdanisur last updated on 31/Aug/21

Find the sum of the real roots of equations: a^3 -6a^2 +15a-17=0 and a^3 -6a^2 +15a-11=0

$$\mathrm{Find}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{real}\:\mathrm{roots}\:\mathrm{of}\:\mathrm{equations}: \\ $$$$\mathrm{a}^{\mathrm{3}} -\mathrm{6a}^{\mathrm{2}} +\mathrm{15a}-\mathrm{17}=\mathrm{0}\:\:\:\mathrm{and} \\ $$$$\mathrm{a}^{\mathrm{3}} -\mathrm{6a}^{\mathrm{2}} +\mathrm{15a}-\mathrm{11}=\mathrm{0} \\ $$

Answered by mr W last updated on 31/Aug/21

eqn. 1: say root is a=2+t (2+t)^3 −6(2+t)^2 +15(2+t)−17=0 ⇒t^3 +3t−3=0 ...(i) eqn. 2: say root is b=2+s (2+s)^3 −6(2+s)^2 +15(2+s)−11=0 ⇒s^3 +3s+3=0 ⇒(−s)^3 +3(−s)−3=0 ...(ii) from (i) and (ii): t=−s sum of real roots of eqn. 1 and 2: a+b=(2+t)+(2+s)=4+t−t=4

$${eqn}.\:\mathrm{1}: \\ $$$${say}\:{root}\:{is}\:{a}=\mathrm{2}+{t} \\ $$$$\left(\mathrm{2}+{t}\right)^{\mathrm{3}} −\mathrm{6}\left(\mathrm{2}+{t}\right)^{\mathrm{2}} +\mathrm{15}\left(\mathrm{2}+{t}\right)−\mathrm{17}=\mathrm{0} \\ $$$$\Rightarrow{t}^{\mathrm{3}} +\mathrm{3}{t}−\mathrm{3}=\mathrm{0}\:\:\:...\left({i}\right) \\ $$$${eqn}.\:\mathrm{2}: \\ $$$${say}\:{root}\:{is}\:{b}=\mathrm{2}+{s} \\ $$$$\left(\mathrm{2}+{s}\right)^{\mathrm{3}} −\mathrm{6}\left(\mathrm{2}+{s}\right)^{\mathrm{2}} +\mathrm{15}\left(\mathrm{2}+{s}\right)−\mathrm{11}=\mathrm{0} \\ $$$$\Rightarrow{s}^{\mathrm{3}} +\mathrm{3}{s}+\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow\left(−{s}\right)^{\mathrm{3}} +\mathrm{3}\left(−{s}\right)−\mathrm{3}=\mathrm{0}\:\:\:...\left({ii}\right) \\ $$$${from}\:\left({i}\right)\:{and}\:\left({ii}\right): \\ $$$${t}=−{s} \\ $$$${sum}\:{of}\:{real}\:{roots}\:{of}\:{eqn}.\:\mathrm{1}\:{and}\:\mathrm{2}: \\ $$$${a}+{b}=\left(\mathrm{2}+{t}\right)+\left(\mathrm{2}+{s}\right)=\mathrm{4}+{t}−{t}=\mathrm{4} \\ $$

Commented by mathdanisur last updated on 01/Sep/21

Thank you Ser

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{Ser} \\ $$

Answered by MJS_new last updated on 31/Aug/21

(1) a=2+u^(1/3) +v^(1/3) (2) a=2−u^(1/3) −v^(1/3) ⇒ answer is 4

$$\left(\mathrm{1}\right)\:{a}=\mathrm{2}+{u}^{\mathrm{1}/\mathrm{3}} +{v}^{\mathrm{1}/\mathrm{3}} \\ $$$$\left(\mathrm{2}\right)\:{a}=\mathrm{2}−{u}^{\mathrm{1}/\mathrm{3}} −{v}^{\mathrm{1}/\mathrm{3}} \\ $$$$\Rightarrow\:\mathrm{answer}\:\mathrm{is}\:\mathrm{4} \\ $$

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