if-and-are-the-roots-of-4x-2-6x-1-00-0-find-3-3-

Question Number 63399 by smartsmith459@gmail.com last updated on 03/Jul/19

i f α a n d β a r e t h e r o o t s o f 4 x^{2} - 6 x + 1 === 00 ======================

= 0 . f i n d α^{3} - β^{3} .

Commented by mr W last updated on 03/Jul/19

α + β = \frac{6}{4} = \frac{3}{2}

α β = \frac{1}{4}

{(α - β)}^{2} = {(α + β)}^{2} - 4 α β

\Rightarrow (α - β) = \pm \sqrt{{(α + β)}^{2} - 4 α β}

α^{3} - β^{3}

= (α - β) (α^{2} + β^{2} + α β)

= (α - β) [{(α + β)}^{2} - α β]

= \pm \sqrt{{(α + β)}^{2} - 4 α β} [{(α + β)}^{2} - α β]

= \pm \sqrt{\frac{9}{4} - 1} [\frac{9}{4} - \frac{1}{4}]

= \pm 2 \sqrt{\frac{5}{4}}

= \pm \sqrt{5}

Answered by MJS last updated on 03/Jul/19

4 x^{2} - 6 x + 1 = 0 \Rightarrow x = \frac{3}{4} \pm \frac{\sqrt{5}}{4}

if α < β \Rightarrow α = \frac{3}{4} - \frac{\sqrt{5}}{4} \land β = \frac{3}{4} + \frac{\sqrt{5}}{4}

\Rightarrow α^{3} - β^{3} = - \sqrt{5}

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