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Let-f-x-x-3-3x-b-and-g-x-x-2-bx-3-where-b-is-a-real-number-What-is-the-sum-of-all-possible-values-of-b-for-which-the-equations-f-x-0-and-g-x-0-have-a-common-root-

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Question Number 19698 by Tinkutara last updated on 14/Aug/17
Let f(x) = x^3 − 3x + b and g(x) = x^2 + bx − 3, where b is a real number. What is the sum of all possible values of b for which the equations f(x) = 0 and g(x) = 0 have a common root?
$$\mathrm{Let}\:{f}\left({x}\right)\:=\:{x}^{\mathrm{3}} \:−\:\mathrm{3}{x}\:+\:{b}\:\mathrm{and}\:{g}\left({x}\right)\:=\:{x}^{\mathrm{2}} \:+ \\ $$$${bx}\:−\:\mathrm{3},\:\mathrm{where}\:{b}\:\mathrm{is}\:\mathrm{a}\:\mathrm{real}\:\mathrm{number}.\:\mathrm{What} \\ $$$$\mathrm{is}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{all}\:\mathrm{possible}\:\mathrm{values}\:\mathrm{of}\:{b}\:\mathrm{for} \\ $$$$\mathrm{which}\:\mathrm{the}\:\mathrm{equations}\:{f}\left({x}\right)\:=\:\mathrm{0}\:\mathrm{and}\:{g}\left({x}\right) \\ $$$$=\:\mathrm{0}\:\mathrm{have}\:\mathrm{a}\:\mathrm{common}\:\mathrm{root}? \\ $$
Answered by ajfour last updated on 14/Aug/17
let α be the common root. f(α)= α^3 −3α+b=0 and αg(α)= α(α^2 +αb−3)=0 subtracting the two we get: α^2 b+3α−3α−b=0 ⇒ α^2 =1 or α=±1 as α^2 +αb−3=0 for α=1, 1+b−3=0 ⇒ b=2 for α=−1, 1−b−3=0 ⇒ b=−2 Σb_i =0 .
$$\mathrm{let}\:\alpha\:\mathrm{be}\:\mathrm{the}\:\mathrm{common}\:\mathrm{root}. \\ $$$$\:\:\:\:\:\:\:\mathrm{f}\left(\alpha\right)=\:\alpha^{\mathrm{3}} −\mathrm{3}\alpha+\mathrm{b}=\mathrm{0} \\ $$$$\mathrm{and}\:\:\:\alpha\mathrm{g}\left(\alpha\right)=\:\alpha\left(\alpha^{\mathrm{2}} +\alpha\mathrm{b}−\mathrm{3}\right)=\mathrm{0} \\ $$$$\mathrm{subtracting}\:\mathrm{the}\:\mathrm{two}\:\mathrm{we}\:\mathrm{get}: \\ $$$$\alpha^{\mathrm{2}} \mathrm{b}+\mathrm{3}\alpha−\mathrm{3}\alpha−\mathrm{b}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\alpha^{\mathrm{2}} =\mathrm{1}\:\:\:\mathrm{or}\:\:\:\:\alpha=\pm\mathrm{1} \\ $$$$\:\:\:\mathrm{as}\:\:\alpha^{\mathrm{2}} +\alpha\mathrm{b}−\mathrm{3}=\mathrm{0} \\ $$$$\:\mathrm{for}\:\alpha=\mathrm{1},\:\:\:\mathrm{1}+\mathrm{b}−\mathrm{3}=\mathrm{0}\:\:\Rightarrow\:\mathrm{b}=\mathrm{2} \\ $$$$\mathrm{for}\:\alpha=−\mathrm{1},\:\:\mathrm{1}−\mathrm{b}−\mathrm{3}=\mathrm{0}\:\:\Rightarrow\:\mathrm{b}=−\mathrm{2} \\ $$$$\:\:\Sigma\mathrm{b}_{\mathrm{i}} =\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 15/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

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