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


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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 byPRITHWISH 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

	$x^{2} - 10 x + 2 = 0$ $α + β = 10 and α β = 2$ $(i) \frac{1}{β} - \frac{1}{α} = \frac{α - β}{α β}$ $but α - β = \sqrt{{(α + β)}^{2} - 4 α β} = \sqrt{100 - 8} = \sqrt{92}$ $\Rightarrow \frac{1}{β} - \frac{1}{α} = \frac{\sqrt{92}}{2}$ $(ii) α^{3} - β^{3} = (α - β) (α^{2} + α β + β^{2})$ $= (α - β) [{(α + β)}^{2} - 2 α β + α β]$ $= \sqrt{92} (100 - 2) = 98 \sqrt{92}$ $\Rightarrow α^{3} - β^{3} = 98 \sqrt{92}$
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