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Question Number 147941 by mathdanisur last updated on 24/Jul/21

$x^{2} - 6 x + 3 = 0 \Rightarrow \frac{2 x^{2}}{3} + \frac{6}{x^{2}} = ?$
	Answered by Rasheed.Sindhi last updated on 24/Jul/21

	$x^{2} - 6 x + 3 = 0 \Rightarrow x^{2} = 6 x - 3$ $\frac{2 x^{2}}{3} + \frac{6}{x^{2}} = \frac{2 (6 x - 3)}{3} + \frac{6}{6 x - 3}$ $= 4 x - 2 + \frac{2}{2 x - 1} = 2 (2 x - 1 + \frac{1}{2 x - 1})$ $x^{2} - 6 x + 3 = 0 \Rightarrow x = \frac{6 \pm \sqrt{36 - 12}}{2}$ $x = 3 \pm \sqrt{6} \Rightarrow 2 x - 1 = 5 \pm 2 \sqrt{6}$ $\frac{2 x^{2}}{3} + \frac{6}{x^{2}} = 2 (2 x - 1 + \frac{1}{2 x - 1})$ $= 2 (5 \pm 2 \sqrt{6} + \frac{1}{5 \pm 2 \sqrt{6}})$ $\frac{1}{5 \pm 2 \sqrt{6}} = \frac{1}{5 \pm 2 \sqrt{6}} \times \frac{5 \mp 2 \sqrt{6}}{5 \mp 2 \sqrt{6}} = \frac{5 \mp 2 \sqrt{6}}{25 - 24} = 5 \mp 2 \sqrt{6}$ $2 (5 \pm 2 \sqrt{6} + 5 \mp 2 \sqrt{6}) = 20$
	Commented by mathdanisur last updated on 24/Jul/21

	$T h a n k y o u S e r$
	Answered by lyubita last updated on 25/Jul/21

	$x^{2} + 3 = 6 x$ $x + \frac{3}{x} = 6$ $x^{2} + \frac{9}{x^{2}} = 30$ $\frac{2 x^{2}}{3} + \frac{6}{x^{2}} = 20$
	Commented by Rasheed.Sindhi last updated on 25/Jul/21

	$N i c e!$
	Commented by lyubita last updated on 25/Jul/21

	$t h a n k y o u m r r a s h e e d$
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