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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 byPRITHWISH 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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