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Question Number 38012 by Rio Mike last updated on 20/Jun/18
The roots of the equation 2x^2 − x + 3 = 0 are α and β if the roots of 3x^2 + px + q=0 are α + (1/α) and β + (1/(β )) find the value of p and q.
$${The}\:{roots}\:{of}\:{the}\:{equation} \\ $$$$\mathrm{2}{x}^{\mathrm{2}} \:−\:{x}\:+\:\mathrm{3}\:=\:\mathrm{0}\:{are}\:\alpha\:{and}\:\beta \\ $$$${if}\:{the}\:{roots}\:{of}\:\mathrm{3}{x}^{\mathrm{2}} \:+\:{px}\:+\:{q}=\mathrm{0}\: \\ $$$${are}\:\alpha\:+\:\frac{\mathrm{1}}{\alpha}\:{and}\:\beta\:+\:\frac{\mathrm{1}}{\beta\:}\:{find}\:{the}\:{value} \\ $$$${of}\:{p}\:{and}\:{q}. \\ $$$$\: \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18
spend time to read book...deaddict from virtual school....
$${spend}\:{time}\:{to}\:{read}\:{book}…{deaddict}\:{from} \\ $$$${virtual}\:{school}…. \\ $$
Answered by MJS last updated on 20/Jun/18
2x^2 −x+3=0 x=(1/4)±(√((1/(16))−(3/2)))=(1/4)±((√(23))/4)i α=(1/4)−((√(23))/4)i ⇒ (1/α)=(1/6)+((√(23))/6)i ⇒ α+(1/α)=(5/(12))−((√(23))/(12))i β=(1/4)+((√(23))/4)i ⇒ (1/β)=(1/6)−((√(23))/6)i ⇒ β+(1/β)=(5/(12))+((√(23))/(12))i 3(x−(5/(12))+((√(23))/(12))i)(x−(5/(12))−((√(23))/(12))i)= =3x^2 −(5/2)x+1 p=−(5/2) q=1
$$\mathrm{2}{x}^{\mathrm{2}} −{x}+\mathrm{3}=\mathrm{0} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{4}}\pm\sqrt{\frac{\mathrm{1}}{\mathrm{16}}−\frac{\mathrm{3}}{\mathrm{2}}}=\frac{\mathrm{1}}{\mathrm{4}}\pm\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}\mathrm{i} \\ $$$$\alpha=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}\mathrm{i}\:\Rightarrow\:\frac{\mathrm{1}}{\alpha}=\frac{\mathrm{1}}{\mathrm{6}}+\frac{\sqrt{\mathrm{23}}}{\mathrm{6}}\mathrm{i}\:\Rightarrow\:\alpha+\frac{\mathrm{1}}{\alpha}=\frac{\mathrm{5}}{\mathrm{12}}−\frac{\sqrt{\mathrm{23}}}{\mathrm{12}}\mathrm{i} \\ $$$$\beta=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\sqrt{\mathrm{23}}}{\mathrm{4}}\mathrm{i}\:\Rightarrow\:\frac{\mathrm{1}}{\beta}=\frac{\mathrm{1}}{\mathrm{6}}−\frac{\sqrt{\mathrm{23}}}{\mathrm{6}}\mathrm{i}\:\Rightarrow\:\beta+\frac{\mathrm{1}}{\beta}=\frac{\mathrm{5}}{\mathrm{12}}+\frac{\sqrt{\mathrm{23}}}{\mathrm{12}}\mathrm{i} \\ $$$$\mathrm{3}\left({x}−\frac{\mathrm{5}}{\mathrm{12}}+\frac{\sqrt{\mathrm{23}}}{\mathrm{12}}\mathrm{i}\right)\left({x}−\frac{\mathrm{5}}{\mathrm{12}}−\frac{\sqrt{\mathrm{23}}}{\mathrm{12}}\mathrm{i}\right)= \\ $$$$=\mathrm{3}{x}^{\mathrm{2}} −\frac{\mathrm{5}}{\mathrm{2}}{x}+\mathrm{1} \\ $$$${p}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${q}=\mathrm{1} \\ $$
Answered by ajfour last updated on 20/Jun/18
αβ = (3/2) , α+β = (1/2) (α+(1/α))(β+(1/β))=(q/3) α+β+(1/α)+(1/β) = −(p/3) ......... ⇒ αβ+(α/β)+(β/α)+(1/(αβ)) = (q/3) (α+β)+((α+β)/(αβ)) = −(p/3) (3/2)+((((1/2))^2 −2((3/2)))/(3/2))+(2/3) = (q/3) and (1/2)+((((1/2)))/(((3/2)))) = −(p/3) q=3[(3/2)−((11)/6)+(2/3)] = 1 p=−3[(1/2)+(1/3)] = −(5/2) .
$$\alpha\beta\:=\:\frac{\mathrm{3}}{\mathrm{2}}\:,\:\:\:\alpha+\beta\:=\:\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\left(\alpha+\frac{\mathrm{1}}{\alpha}\right)\left(\beta+\frac{\mathrm{1}}{\beta}\right)=\frac{{q}}{\mathrm{3}} \\ $$$$\alpha+\beta+\frac{\mathrm{1}}{\alpha}+\frac{\mathrm{1}}{\beta}\:=\:−\frac{{p}}{\mathrm{3}} \\ $$$$……… \\ $$$$\Rightarrow\:\:\:\alpha\beta+\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+\frac{\mathrm{1}}{\alpha\beta}\:=\:\frac{{q}}{\mathrm{3}} \\ $$$$\:\:\:\:\:\:\left(\alpha+\beta\right)+\frac{\alpha+\beta}{\alpha\beta}\:=\:−\frac{{p}}{\mathrm{3}} \\ $$$$\:\:\:\:\frac{\mathrm{3}}{\mathrm{2}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\mathrm{2}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}{\frac{\mathrm{3}}{\mathrm{2}}}+\frac{\mathrm{2}}{\mathrm{3}}\:=\:\frac{{q}}{\mathrm{3}} \\ $$$${and}\:\:\:\frac{\mathrm{1}}{\mathrm{2}}+\frac{\left(\frac{\mathrm{1}}{\mathrm{2}}\right)}{\left(\frac{\mathrm{3}}{\mathrm{2}}\right)}\:=\:−\frac{{p}}{\mathrm{3}} \\ $$$${q}=\mathrm{3}\left[\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{11}}{\mathrm{6}}+\frac{\mathrm{2}}{\mathrm{3}}\right]\:=\:\mathrm{1} \\ $$$${p}=−\mathrm{3}\left[\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{3}}\right]\:=\:−\frac{\mathrm{5}}{\mathrm{2}}\:. \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 20/Jun/18
α+β=(1/(2 )) αβ=(3/2) α+(1/α)+β+(1/β)=((−p)/3) α+β+((α+β)/(αβ))=((−p)/3) (1/2)+(1/2)×(2/3)=−(p/3) (5/6)=−(p/3) p=−(5/2) (α+(1/α))×(β+(1/β))=(q/3) αβ+(α/β)+(β/α)+(1/(αβ))=(q/3) αβ+(1/(αβ))+((α^2 +β^2 )/(αβ))=(q/3) (3/2)+(2/3)+(((α+β)^2 −2αβ)/(αβ))=(q/3) ((13)/6)+(((1/4)−2×(3/2))/(3/2))=(q/3) ((13)/6)+(((1−12)/4)/(3/2))=(q/3) ((13)/6)−((11×2)/(12))=(q/3) ((13−11)/6)=(q/3) q=1
$$\alpha+\beta=\frac{\mathrm{1}}{\mathrm{2}\:\:}\:\:\alpha\beta=\frac{\mathrm{3}}{\mathrm{2}} \\ $$$$\alpha+\frac{\mathrm{1}}{\alpha}+\beta+\frac{\mathrm{1}}{\beta}=\frac{−{p}}{\mathrm{3}} \\ $$$$\alpha+\beta+\frac{\alpha+\beta}{\alpha\beta}=\frac{−{p}}{\mathrm{3}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}×\frac{\mathrm{2}}{\mathrm{3}}=−\frac{{p}}{\mathrm{3}} \\ $$$$\frac{\mathrm{5}}{\mathrm{6}}=−\frac{{p}}{\mathrm{3}} \\ $$$${p}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$$\left(\alpha+\frac{\mathrm{1}}{\alpha}\right)×\left(\beta+\frac{\mathrm{1}}{\beta}\right)=\frac{{q}}{\mathrm{3}} \\ $$$$\alpha\beta+\frac{\alpha}{\beta}+\frac{\beta}{\alpha}+\frac{\mathrm{1}}{\alpha\beta}=\frac{{q}}{\mathrm{3}} \\ $$$$\alpha\beta+\frac{\mathrm{1}}{\alpha\beta}+\frac{\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} }{\alpha\beta}=\frac{{q}}{\mathrm{3}} \\ $$$$\frac{\mathrm{3}}{\mathrm{2}}+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta}{\alpha\beta}=\frac{{q}}{\mathrm{3}} \\ $$$$\frac{\mathrm{13}}{\mathrm{6}}+\frac{\frac{\mathrm{1}}{\mathrm{4}}−\mathrm{2}×\frac{\mathrm{3}}{\mathrm{2}}}{\frac{\mathrm{3}}{\mathrm{2}}}=\frac{{q}}{\mathrm{3}} \\ $$$$\frac{\mathrm{13}}{\mathrm{6}}+\frac{\frac{\mathrm{1}−\mathrm{12}}{\mathrm{4}}}{\frac{\mathrm{3}}{\mathrm{2}}}=\frac{{q}}{\mathrm{3}} \\ $$$$\frac{\mathrm{13}}{\mathrm{6}}−\frac{\mathrm{11}×\mathrm{2}}{\mathrm{12}}=\frac{{q}}{\mathrm{3}} \\ $$$$\frac{\mathrm{13}−\mathrm{11}}{\mathrm{6}}=\frac{{q}}{\mathrm{3}} \\ $$$${q}=\mathrm{1} \\ $$

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