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Question Number 166195 by mathlove last updated on 15/Feb/22

	Answered by alephzero last updated on 15/Feb/22

	$x = 1$
	Commented by mathlove last updated on 15/Feb/22

	$h o w ?$
	Commented by alephzero last updated on 15/Feb/22


	Commented by alephzero last updated on 15/Feb/22

	$t h e g r a p h i n t e r s e c t s t h e x - a x i s$ $a t p o i n t 1$
	Commented by mr W last updated on 15/Feb/22

	$i {c a n}^{'} t s e e t h a t y o u r g r a p h i n t e r s e c t s$ $t h e x - a x i s a t a l l . y o u j u s t i m a g e t h a t$ $i t i n t e r s e c t s x - a x i s a t 1 . t h i s {i s n}^{'} t$ $a s t r i c t p r o o f t h a t x = 1 i s a r o o t a n d$ $t h e o n l y r o o t .$
	Answered by mathsmine last updated on 15/Feb/22

	$x (\sqrt{x} - 1) ⩾ 0$ $\Rightarrow x \in [1, + \infty [$ $\Leftrightarrow 1 - x = \sqrt{x \sqrt{x} - x}$ $i f x > 1 \Rightarrow 1 - x < 0 w e g e t \sqrt{x \sqrt{x} - x} < 0$ $i f x = 1 \Rightarrow 0 = 0 t r u e$ $S = {1}$
	Answered by TheSupreme last updated on 15/Feb/22
	$x≥0 x((√x)−1)≥0 →x≥1 x=1 →f(x)=0 f′(x≥1)=1+(((3/2)x^(1/2) −1)/(2(√(x(√x)−1 ))))≥1+(0/(2(√(x(√x)−1))))≥1 ∀x∈{[1,∞) so x=1 is the only one solution$
	$x ⩾ 0$ $x (\sqrt{x} - 1) ⩾ 0 \to x ⩾ 1$ $x = 1 \to f (x) = 0$ $f^{'} (x ⩾ 1) = 1 + \frac{\frac{3}{2} x^{\frac{1}{2}} - 1}{2 \sqrt{x \sqrt{x} - 1}} ⩾ 1 + \frac{0}{2 \sqrt{x \sqrt{x} - 1}} ⩾ 1 \forall x \in {[1, \infty)$ $s o x = 1 i s t h e o n l y o n e s o l u t i o n$
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