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Question Number 140981 by EnterUsername last updated on 14/May/21

Let α≠1 and α^(13) =1. If a=α+α^3 +α^4 +α^(−4) +α^(−3) + α^(−1) and b=α^2 +α^5 +α^6 +α^(−6) +α^(−5) +α^(−2) then the quadratic equation whose roots are a and b is (A) x^2 +x+3=0 (B) x^2 +x+4=0 (C) x^2 +x−3=0 (D) x^2 +x−4=0

$$\mathrm{Let}\:\alpha\neq\mathrm{1}\:\mathrm{and}\:\alpha^{\mathrm{13}} =\mathrm{1}.\:\mathrm{If}\:{a}=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{−\mathrm{4}} +\alpha^{−\mathrm{3}} + \\ $$$$\alpha^{−\mathrm{1}} \:\mathrm{and}\:{b}=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{−\mathrm{6}} +\alpha^{−\mathrm{5}} +\alpha^{−\mathrm{2}} \:\mathrm{then}\:\mathrm{the} \\ $$$$\mathrm{quadratic}\:\mathrm{equation}\:\mathrm{whose}\:\mathrm{roots}\:\mathrm{are}\:{a}\:\mathrm{and}\:{b}\:\mathrm{is} \\ $$$$\left(\mathrm{A}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{B}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}+\mathrm{4}=\mathrm{0} \\ $$$$\left(\mathrm{C}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{3}=\mathrm{0}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\mathrm{D}\right)\:\mathrm{x}^{\mathrm{2}} +\mathrm{x}−\mathrm{4}=\mathrm{0} \\ $$

Answered by Rasheed.Sindhi last updated on 16/May/21

FORMULAE: For nth roots α^(k+n) =α^k 1+α+α^2 +α^3 +...+α^n =0 (Sum of all the nth roots is zero) α+α^2 +α^3 +...+α^n =−1 For 13th roots α^(k+13) =α^k 1+α+α^2 +α^3 +...+α^(12) =0 (Sum of all the 13th roots is zero) α+α^2 +α^3 +...+α^(12) =−1 ⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌣⌢⌣⌢ a=α+α^3 +α^4 +α^(−4) +α^(−3) +α^(−1) =α+α^3 +α^4 +α^9 +α^(10) +α^(12) b=α^2 +α^5 +α^6 +α^(−6) +α^(−5) +α^(−2) =α^2 +α^5 +α^6 +α^7 +α^8 +α^(11) a+b=α+α^2 +α^3 +...+α^(12) =−1 ab=(α+α^3 +α^4 +α^9 +α^(10) +α^(12) ) ×(α^2 +α^5 +α^6 +α^7 +α^8 +α^(11) ) =α^3 +α^6 +α^7 +α^8 +α^9 +α^(12) +α^5 +α^8 +α^9 +α^(10) +α^(11) +α^1 +α^6 +α^9 +α^(10) +α^(11) +α^(12) +α^2 +α^(11) +α^1 +α^2 +α^3 +α^4 +α^7 +α^(12) +α^2 +α^3 +α^4 +α^5 +α^8 +α^1 +α^4 +α^5 +α^6 +α^7 +α^(10) =(α^1 +α^2 +α^3 +...+α^(12) ) +(α^1 +α^2 +α^3 +...+α^(12) ) +(α^1 +α^2 +α^3 +...+α^(12) ) =(−1)+(−1)+(−1)=−3 Equation required: x^2 −(a+b)x+ab x^2 −(−1)x+(−3) x^2 +x−3 So (C) is correct

$${FORMULAE}: \\ $$$$\mathcal{F}{or}\:{nth}\:{roots} \\ $$$$\alpha^{{k}+{n}} =\alpha^{{k}} \\ $$$$\mathrm{1}+\alpha+\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{{n}} =\mathrm{0} \\ $$$$\left({Sum}\:{of}\:{all}\:{the}\:{nth}\:{roots}\:{is}\:{zero}\right) \\ $$$$\alpha+\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{{n}} =−\mathrm{1} \\ $$$$\mathcal{F}{or}\:\mathrm{13}{th}\:{roots} \\ $$$$\:\:\:\:\alpha^{{k}+\mathrm{13}} =\alpha^{{k}} \\ $$$$\mathrm{1}+\alpha+\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{\mathrm{12}} =\mathrm{0} \\ $$$$\left({Sum}\:{of}\:{all}\:{the}\:\mathrm{13}{th}\:{roots}\:{is}\:{zero}\right) \\ $$$$\alpha+\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{\mathrm{12}} =−\mathrm{1} \\ $$$$\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\frown\smile\smile\frown\smile\frown \\ $$$${a}=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{−\mathrm{4}} +\alpha^{−\mathrm{3}} +\alpha^{−\mathrm{1}} \\ $$$$\:\:\:=\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{9}} +\alpha^{\mathrm{10}} +\alpha^{\mathrm{12}} \\ $$$${b}=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{−\mathrm{6}} +\alpha^{−\mathrm{5}} +\alpha^{−\mathrm{2}} \\ $$$$\:\:\:=\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{\mathrm{7}} +\alpha^{\mathrm{8}} +\alpha^{\mathrm{11}} \\ $$$${a}+{b}=\alpha+\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{\mathrm{12}} =−\mathrm{1} \\ $$$${ab}=\left(\alpha+\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{9}} +\alpha^{\mathrm{10}} +\alpha^{\mathrm{12}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:×\left(\alpha^{\mathrm{2}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{\mathrm{7}} +\alpha^{\mathrm{8}} +\alpha^{\mathrm{11}} \right) \\ $$$$\:\:\:\:=\alpha^{\mathrm{3}} +\alpha^{\mathrm{6}} +\alpha^{\mathrm{7}} +\alpha^{\mathrm{8}} +\alpha^{\mathrm{9}} +\alpha^{\mathrm{12}} \\ $$$$\:\:\:\:\:\:\:+\alpha^{\mathrm{5}} +\alpha^{\mathrm{8}} +\alpha^{\mathrm{9}} +\alpha^{\mathrm{10}} +\alpha^{\mathrm{11}} +\alpha^{\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:+\alpha^{\mathrm{6}} +\alpha^{\mathrm{9}} +\alpha^{\mathrm{10}} +\alpha^{\mathrm{11}} +\alpha^{\mathrm{12}} +\alpha^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:+\alpha^{\mathrm{11}} +\alpha^{\mathrm{1}} +\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{7}} \\ $$$$\:\:\:\:\:\:\:+\alpha^{\mathrm{12}} +\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{8}} \\ $$$$\:\:\:\:\:\:\:+\alpha^{\mathrm{1}} +\alpha^{\mathrm{4}} +\alpha^{\mathrm{5}} +\alpha^{\mathrm{6}} +\alpha^{\mathrm{7}} +\alpha^{\mathrm{10}} \\ $$$$\:\:=\left(\alpha^{\mathrm{1}} +\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{\mathrm{12}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:+\left(\alpha^{\mathrm{1}} +\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{\mathrm{12}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\left(\alpha^{\mathrm{1}} +\alpha^{\mathrm{2}} +\alpha^{\mathrm{3}} +...+\alpha^{\mathrm{12}} \right) \\ $$$$\:=\left(−\mathrm{1}\right)+\left(−\mathrm{1}\right)+\left(−\mathrm{1}\right)=−\mathrm{3} \\ $$$$\mathcal{E}{quation}\:{required}: \\ $$$$\:\:\:\:\:{x}^{\mathrm{2}} −\left({a}+{b}\right){x}+{ab} \\ $$$$\:\:\:\:{x}^{\mathrm{2}} −\left(−\mathrm{1}\right){x}+\left(−\mathrm{3}\right) \\ $$$$\:\:\:\:{x}^{\mathrm{2}} +{x}−\mathrm{3} \\ $$$${So}\:\left({C}\right)\:{is}\:{correct} \\ $$

Commented by EnterUsername last updated on 19/May/21

Thank you Sir

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

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