If x^2 − 65x = 64(√x) then (√(x − (√x) )) = ?


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Question Number 194226 by BaliramKumar last updated on 30/Jun/23

$If x^{2} - 65 x = 64 \sqrt{x} then \sqrt{x - \sqrt{x}} = ?$
	Answered by Frix last updated on 30/Jun/23

	$x^{2} - 65 x = 64 \sqrt{x}$ $x = 0 \Rightarrow \sqrt{x - \sqrt{x}} = 0 ★$ $x^{\frac{3}{2}} - 65 \sqrt{x} - 64 = 0$ $x = t^{2} \land t > 0$ $t^{3} - 65 t - 64 = 0$ $(t + 1) (t^{2} - t - 64) = 0$ $t = \frac{1 + \sqrt{257}}{2} = \sqrt{x}$ $x = \frac{129 + \sqrt{257}}{2}$ $\Rightarrow \sqrt{x - \sqrt{x}} = 8 ★$
	Answered by Frix last updated on 30/Jun/23

	$x^{2} - 65 x - 64 \sqrt{x} = 0$ $\sqrt{x} (\sqrt{x} + 1) (x - \sqrt{x} - 64) = 0$ $x = 0 ★$ $\sqrt{x - \sqrt{x}} = 0$ $x - \sqrt{x} - 64 = 0$ $x - \sqrt{x} = 64$ $\sqrt{x - \sqrt{x}} = 8 ★$
	Answered by manxsol last updated on 30/Jun/23

	$x^{2} - x = 64 (x + \sqrt{x})$ ${(\sqrt{x})}^{4} - {(\sqrt{x})}^{2} = 64 (x + \sqrt{x})$ $(\sqrt{x}^{2} - \sqrt{x}) (\sqrt{x}^{2} + \sqrt{x}) = 64 (x + \sqrt{x})$ $(x - \sqrt{x}) (x + \sqrt{x}) = 64 (x + \sqrt{x})$ $x - \sqrt{x} = 64$ $\sqrt{x - \sqrt{x}} = 8 s o l I$ $x + \sqrt{x} = 0$ $\sqrt{x} (\sqrt{x} + 1) = 0$ $x = 0 \sqrt{x - \sqrt{x}} = 0 s o l I I$ $\sqrt{x} = - 1 \times$
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