solve x^2 +(√(3−x))=3


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Question Number 170844 by mr W last updated on 01/Jun/22

$s o l v e x^{2} + \sqrt{3 - x} = 3$
	Answered by balirampatel last updated on 01/Jun/22

	$s o l u t i o n :$ $x^{2} + \sqrt{3 - x} = 3$ $x^{2} = 3 - \sqrt{3 - x}$ $x = \sqrt{3 - \sqrt{3 - x}}$ $x = \sqrt{3 - \sqrt{3 - \sqrt{3 - \sqrt{3 - x}}}}$ $x = \sqrt{3 - \sqrt{3 - \sqrt{3 - . . . . . . . . \infty}}}$ $x = \sqrt{3 - x}$ $x^{2} = 3 - x$ $x^{2} + x - 3 = 0$ $x = \frac{- 1 \pm \sqrt{1^{2} - 4 \cdot 1 \cdot (- 3)}}{2 \cdot 1}$ $x = \frac{- 1 \pm \sqrt{1 + 12}}{2}$ $x = \frac{- 1 \pm \sqrt{13}}{2}$ $x = \frac{- 1 + \sqrt{13}}{2} A n s w e r [o n l y p o s i t i v e v a l u e s a t i s f i e d]$ $[a l s o x = - 1]$
	Commented by mr W last updated on 01/Jun/22

	$n i c e s o l u t i o n!$
	Answered by MJS_new last updated on 01/Jun/22

	$x^{2} + \sqrt{3 - x} = 3$ $\sqrt{3 - x} = t ⩾ 0$ $t^{4} - 6 t^{2} + t + 6 = 0$ $(t - 2) (t + 1) (t^{2} + t - 3) = 0$ $t ⩾ 0 \Rightarrow t = \frac{- 1 + \sqrt{13}}{2} \lor t = 2$ $\Rightarrow x = \frac{- 1 + \sqrt{13}}{2} \lor x = - 1$
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