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

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

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

$$FORMULAE:$$$$\mathcal{F}ornthroots$$$${\alpha }^{k+n}={\alpha }^k$$$$1+\alpha +{\alpha }^2+{\alpha }^3+...+{\alpha }^n=0$$$$\left(Sumofallthenthrootsiszero\right)$$$$\alpha +{\alpha }^2+{\alpha }^3+...+{\alpha }^n=-1$$$$\mathcal{F}or13throots$$$${\alpha }^{k+13}={\alpha }^k$$$$1+\alpha +{\alpha }^2+{\alpha }^3+...+{\alpha }^{12}=0$$$$\left(Sumofallthe13throotsiszero\right)$$$$\alpha +{\alpha }^2+{\alpha }^3+...+{\alpha }^{12}=-1$$$$⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌢⌣⌣⌢⌣⌢$$$$a=\alpha +{\alpha }^3+{\alpha }^4+{\alpha }^{-4}+{\alpha }^{-3}+{\alpha }^{-1}$$$$=\alpha +{\alpha }^3+{\alpha }^4+{\alpha }^9+{\alpha }^{10}+{\alpha }^{12}$$$$b={\alpha }^2+{\alpha }^5+{\alpha }^6+{\alpha }^{-6}+{\alpha }^{-5}+{\alpha }^{-2}$$$$={\alpha }^2+{\alpha }^5+{\alpha }^6+{\alpha }^7+{\alpha }^8+{\alpha }^{11}$$$$a+b=\alpha +{\alpha }^2+{\alpha }^3+...+{\alpha }^{12}=-1$$$$ab=(\alpha +{\alpha }^3+{\alpha }^4+{\alpha }^9+{\alpha }^{10}+{\alpha }^{12})$$$$\times ({\alpha }^2+{\alpha }^5+{\alpha }^6+{\alpha }^7+{\alpha }^8+{\alpha }^{11})$$$$={\alpha }^3+{\alpha }^6+{\alpha }^7+{\alpha }^8+{\alpha }^9+{\alpha }^{12}$$$$+{\alpha }^5+{\alpha }^8+{\alpha }^9+{\alpha }^{10}+{\alpha }^{11}+{\alpha }^1$$$$+{\alpha }^6+{\alpha }^9+{\alpha }^{10}+{\alpha }^{11}+{\alpha }^{12}+{\alpha }^2$$$$+{\alpha }^{11}+{\alpha }^1+{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^7$$$$+{\alpha }^{12}+{\alpha }^2+{\alpha }^3+{\alpha }^4+{\alpha }^5+{\alpha }^8$$$$+{\alpha }^1+{\alpha }^4+{\alpha }^5+{\alpha }^6+{\alpha }^7+{\alpha }^{10}$$$$=({\alpha }^1+{\alpha }^2+{\alpha }^3+...+{\alpha }^{12})$$$$+({\alpha }^1+{\alpha }^2+{\alpha }^3+...+{\alpha }^{12})$$$$+({\alpha }^1+{\alpha }^2+{\alpha }^3+...+{\alpha }^{12})$$$$=(-1)+(-1)+(-1)=-3$$$$\mathcal{E}quationrequired:$$$$x^2-(a+b)x+ab$$$$x^2-(-1)x+(-3)$$$$x^2+x-3$$$$So\left(C\right)iscorrect$$

Commented by EnterUsername last updated on 19/May/21

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

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