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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} - 10 x + 2 = 0 and α > β, find :

(i) \frac{1}{β} - \frac{1}{α}

(ii) α^{3} - β^{3}

Commented by PRITHWISH SEN 2 last updated on 01/Oct/20

putting y = \frac{1}{x}

{2 y}^{2} - 10 y + 1 = 0

diff . of root

\frac{1}{β} - \frac{1}{α} = \frac{\sqrt{D}}{a} = \frac{\sqrt{92}}{2}

Answered by bemath last updated on 01/Oct/20

(i) \frac{1}{β} - \frac{1}{α} = \frac{α - β}{α β} = \frac{\sqrt{100 - 4.1.2}}{2}

= \frac{\sqrt{92}}{2}

Answered by Dwaipayan Shikari last updated on 01/Oct/20

x^{2} - 10 x + 2 = 0

α^{3} - β^{3} = (α - β) (α^{2} + α β + β^{2}) = \sqrt{92} ({(α + β)}^{2} - α β) = 2 \sqrt{23} (100 - 2) = 196 \sqrt{23}

α + β = 10

α β = 2

α - β = \sqrt{100 - 8} = \sqrt{92}

Answered by Rio Michael last updated on 01/Oct/20

x^{2} - 10 x + 2 = 0

α + β = 10 and α β = 2

(i) \frac{1}{β} - \frac{1}{α} = \frac{α - β}{α β}

but α - β = \sqrt{{(α + β)}^{2} - 4 α β} = \sqrt{100 - 8} = \sqrt{92}

\Rightarrow \frac{1}{β} - \frac{1}{α} = \frac{\sqrt{92}}{2}

(ii) α^{3} - β^{3} = (α - β) (α^{2} + α β + β^{2})

= (α - β) [{(α + β)}^{2} - 2 α β + α β]

= \sqrt{92} (100 - 2) = 98 \sqrt{92}

\Rightarrow α^{3} - β^{3} = 98 \sqrt{92}