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Question Number 25184 by lucky singh last updated on 05/Dec/17
if a and b are the root of the quadratic equation x^2 +2x+3 then find the value of α^2 /β+β^2 /α
$${if}\:{a}\:{and}\:{b}\:{are}\:{the}\:{root}\:{of}\:{the}\:{quadratic}\:{equation}\:{x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}\:{then}\:{find}\:{the}\:{value}\:{of}\:\alpha^{\mathrm{2}} /\beta+\beta^{\mathrm{2}} /\alpha \\ $$
Commented by prakash jain last updated on 05/Dec/17
x^2 +2x+3=0  α+β=−2  αβ=3  (α^2 /β)+(β^2 /α)  =((α^3 +β^3 )/(αβ))=(((α+β)(α^2 +β^2 −αβ))/(αβ))  =(((α+β)((α+β)^2 −3αβ))/(αβ))  =((−2((−2)^2 −3×3))/3)  =((−2(−5))/3)=((10)/3)
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{3}=\mathrm{0} \\ $$$$\alpha+\beta=−\mathrm{2} \\ $$$$\alpha\beta=\mathrm{3} \\ $$$$\frac{\alpha^{\mathrm{2}} }{\beta}+\frac{\beta^{\mathrm{2}} }{\alpha} \\ $$$$=\frac{\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} }{\alpha\beta}=\frac{\left(\alpha+\beta\right)\left(\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} −\alpha\beta\right)}{\alpha\beta} \\ $$$$=\frac{\left(\alpha+\beta\right)\left(\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{3}\alpha\beta\right)}{\alpha\beta} \\ $$$$=\frac{−\mathrm{2}\left(\left(−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{3}×\mathrm{3}\right)}{\mathrm{3}} \\ $$$$=\frac{−\mathrm{2}\left(−\mathrm{5}\right)}{\mathrm{3}}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$
Answered by ibraheem160 last updated on 06/Dec/17
α+β=−2  αβ=3  (α^2 /β)+(β^2 /α)=((α^3 +β^3 )/(αβ))  α^3 +β^3 =(α+β)^3 −3αβ(α+β)  ⇒(−2)^3 −3(3)(−2)  =10  ∴ ((α^3 +β^3 )/(αβ))=((10)/3)
$$\alpha+\beta=−\mathrm{2} \\ $$$$\alpha\beta=\mathrm{3} \\ $$$$\frac{\alpha^{\mathrm{2}} }{\beta}+\frac{\beta^{\mathrm{2}} }{\alpha}=\frac{\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} }{\alpha\beta} \\ $$$$\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} =\left(\alpha+\beta\right)^{\mathrm{3}} −\mathrm{3}\alpha\beta\left(\alpha+\beta\right) \\ $$$$\Rightarrow\left(−\mathrm{2}\right)^{\mathrm{3}} −\mathrm{3}\left(\mathrm{3}\right)\left(−\mathrm{2}\right) \\ $$$$=\mathrm{10} \\ $$$$\therefore\:\frac{\alpha^{\mathrm{3}} +\beta^{\mathrm{3}} }{\alpha\beta}=\frac{\mathrm{10}}{\mathrm{3}} \\ $$

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