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Question Number 116138 by harckinwunmy last updated on 01/Oct/20
If α and β are the roots of the quadratic equation x^2 −10x+2=0 and α >β, find: (i) (1/β)−(1/α) (ii)α^3 −β^3
$$\mathrm{If}\:\:\alpha\:\mathrm{and}\:\beta\:\mathrm{are}\:\mathrm{the}\:\mathrm{roots}\:\mathrm{of}\: \\ $$$$\mathrm{the}\:\mathrm{quadratic}\:\mathrm{equation}\: \\ $$$$\mathrm{x}^{\mathrm{2}} −\mathrm{10x}+\mathrm{2}=\mathrm{0}\:\mathrm{and}\:\alpha\:>\beta,\:\mathrm{find}: \\ $$$$\left(\mathrm{i}\right)\:\frac{\mathrm{1}}{\beta}−\frac{\mathrm{1}}{\alpha} \\ $$$$\left(\mathrm{ii}\right)\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} \\ $$
Commented by PRITHWISH SEN 2 last updated on 01/Oct/20
putting y=(1/x) 2y^2 −10y+1=0 diff. of root (1/β)−(1/α) = ((√D)/a)=((√(92))/2)
$$\mathrm{putting}\:\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{x}} \\ $$$$\mathrm{2y}^{\mathrm{2}} −\mathrm{10y}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{diff}.\:\mathrm{of}\:\mathrm{root} \\ $$$$\frac{\mathrm{1}}{\beta}−\frac{\mathrm{1}}{\alpha}\:=\:\frac{\sqrt{\mathrm{D}}}{\mathrm{a}}=\frac{\sqrt{\mathrm{92}}}{\mathrm{2}} \\ $$
Answered by bemath last updated on 01/Oct/20
(i) (1/β)−(1/α)=((α−β)/(αβ)) = ((√(100−4.1.2))/2) =((√(92))/2)
$$\left(\mathrm{i}\right)\:\frac{\mathrm{1}}{\beta}−\frac{\mathrm{1}}{\alpha}=\frac{\alpha−\beta}{\alpha\beta}\:=\:\frac{\sqrt{\mathrm{100}−\mathrm{4}.\mathrm{1}.\mathrm{2}}}{\mathrm{2}} \\ $$$$\:\:\:\:=\frac{\sqrt{\mathrm{92}}}{\mathrm{2}}\: \\ $$$$ \\ $$
Answered by Dwaipayan Shikari last updated on 01/Oct/20
x^2 −10x+2=0 α^3 −β^3 =(α−β)(α^2 +αβ+β^2 )=(√(92))((α+β)^2 −αβ)=2(√(23)) (100−2)=196(√(23)) α+β=10 αβ=2 α−β=(√(100−8))=(√(92))
$$\mathrm{x}^{\mathrm{2}} −\mathrm{10x}+\mathrm{2}=\mathrm{0} \\ $$$$\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} =\left(\alpha−\beta\right)\left(\alpha^{\mathrm{2}} +\alpha\beta+\beta^{\mathrm{2}} \right)=\sqrt{\mathrm{92}}\left(\left(\alpha+\beta\right)^{\mathrm{2}} −\alpha\beta\right)=\mathrm{2}\sqrt{\mathrm{23}}\:\left(\mathrm{100}−\mathrm{2}\right)=\mathrm{196}\sqrt{\mathrm{23}} \\ $$$$\alpha+\beta=\mathrm{10} \\ $$$$\alpha\beta=\mathrm{2} \\ $$$$\alpha−\beta=\sqrt{\mathrm{100}−\mathrm{8}}=\sqrt{\mathrm{92}} \\ $$
Answered by Rio Michael last updated on 01/Oct/20
x^2 −10x + 2 = 0 α + β = 10 and αβ = 2 (i) (1/β)−(1/α) = ((α−β)/(αβ)) but α−β = (√((α+β)^2 −4αβ)) = (√(100−8)) = (√(92)) ⇒ (1/β)−(1/α) = ((√(92))/2) (ii) α^3 −β^3 = (α−β)(α^2 + αβ + β^2 ) = (α−β)[(α+β)^2 −2αβ + αβ] = (√(92))( 100 − 2) = 98(√(92)) ⇒ α^3 −β^3 = 98(√(92))
$${x}^{\mathrm{2}} −\mathrm{10}{x}\:+\:\mathrm{2}\:=\:\mathrm{0} \\ $$$$\:\alpha\:+\:\beta\:=\:\mathrm{10}\:\mathrm{and}\:\alpha\beta\:=\:\mathrm{2} \\ $$$$\left(\mathrm{i}\right)\:\frac{\mathrm{1}}{\beta}−\frac{\mathrm{1}}{\alpha}\:=\:\frac{\alpha−\beta}{\alpha\beta} \\ $$$$\mathrm{but}\:\alpha−\beta\:=\:\sqrt{\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{4}\alpha\beta}\:=\:\sqrt{\mathrm{100}−\mathrm{8}}\:=\:\sqrt{\mathrm{92}} \\ $$$$\Rightarrow\:\frac{\mathrm{1}}{\beta}−\frac{\mathrm{1}}{\alpha}\:=\:\frac{\sqrt{\mathrm{92}}}{\mathrm{2}} \\ $$$$\left(\mathrm{ii}\right)\:\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} \:=\:\left(\alpha−\beta\right)\left(\alpha^{\mathrm{2}} \:+\:\alpha\beta\:+\:\beta^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\left(\alpha−\beta\right)\left[\left(\alpha+\beta\right)^{\mathrm{2}} −\mathrm{2}\alpha\beta\:\:+\:\alpha\beta\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\sqrt{\mathrm{92}}\left(\:\mathrm{100}\:−\:\mathrm{2}\right)\:=\:\mathrm{98}\sqrt{\mathrm{92}} \\ $$$$\:\Rightarrow\:\alpha^{\mathrm{3}} −\beta^{\mathrm{3}} \:=\:\mathrm{98}\sqrt{\mathrm{92}} \\ $$

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