solve for x∈R (√(x^2 −4))+2(√(x^2 −1))=x^2


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Question Number 173732 by mr W last updated on 17/Jul/22

$s o l v e f o r x \in R$ $\sqrt{x^{2} - 4} + 2 \sqrt{x^{2} - 1} = x^{2}$
Commented by dragan91 last updated on 18/Jul/22

$let x^{2} - 2 = t$ $\sqrt{t - 2} + 2 \sqrt{t + 1} = t + 2 /^{2}$ $t - 2 + 4 \sqrt{t - 2} \sqrt{t + 1} + 4 t + 4 = t^{2} + 4 t + 4$ $4 \sqrt{(t - 2) (t + 1)} = t^{2} - t + 2 /^{2}$ ${16 t}^{2} - 16 t - 32 = t^{4} + t^{2} + 4 - {2 t}^{3} + {4 t}^{2} - 4 t$ $t^{4} - {2 t}^{3} - {11 t}^{2} + 12 t + 36 = 0$ $t^{4} + {4 t}^{3} + {4 t}^{2} - {6 t}^{3} - {12 t}^{2} - {3 t}^{2} - 6 t + 18 t + 36 = 0$ $t^{2} {(t + 2)}^{2} - {6 t}^{2} (t + 2) - 3 t (t + 2) + 18 (t + 2) = 0$ $(t + 2) (t^{3} - {4 t}^{2} - 3 t + 18) = 0$ $(t + 2) (t^{3} - {6 t}^{2} + 9 t + {2 t}^{2} - 12 t + 18) = 0$ $(t + 2) (t {(t - 3)}^{2} + 2 {(t - 3)}^{2}) = 0$ ${(t + 2)}^{2} {(t - 3)}^{2} = 0$ $t = - 2 reject$ $t = 3$ $x^{2} - 2 = 3 \Rightarrow x = \pm \sqrt{5}$
	Answered by Skabetix last updated on 17/Jul/22

	$x = \sqrt{} 5 o r x = - \sqrt{} 5 ? ?$
	Answered by mindispower last updated on 17/Jul/22

	$t = x^{2}$ $\sqrt{t - 4} + 2 \sqrt{t - 1} = t, t ⩾ 4$ $t - 2 \sqrt{t - 1} = {(\sqrt{t - 1} - 1)}^{2}$ $\Leftrightarrow \sqrt{t - 4} = {(\sqrt{t - 1} - 1)}^{2}$ $\Rightarrow \sqrt{t - 4} = \sqrt{t - 1} - 1 > 0$ $\sqrt{t - 4} - \sqrt{t - 1} = - 1$ $\Rightarrow \frac{- 3}{\sqrt{t - 4} + \sqrt{t - 1}} = - 1$ $\Rightarrow \sqrt{t - 4} + \sqrt{t - 1} = 3$ $\Rightarrow 2 \sqrt{t - 4} = 2$ $\Rightarrow t = 5 \Rightarrow x = \underline{+} \sqrt{5}$
	Commented by mr W last updated on 17/Jul/22

	$t h a n k s s i r!$ $b u t h o w d i d y o u g e t ?$ $\sqrt{t - 4} = {(\sqrt{t - 1} - 1)}^{2}$ $\Rightarrow \sqrt{t - 4} = \sqrt{t - 1} - 1$
	Commented by Tawa11 last updated on 17/Jul/22

	$Great sirs$
	Answered by behi834171 last updated on 17/Jul/22
	$x^2 −4=t^2 ⇒t+2(√(t^2 +3))=t^2 +4⇒ 4(t^2 +3)=t^4 +8t^2 +t^2 −2t^3 −8t+16⇒ t^4 −2t^3 +5t^2 −8t+4=0 ⇒(t−1)^2 (t^2 +4)=0⇒t^2 =1,−4 ⇒ { ((x^2 −4=1⇒x^2 =5⇒x=±(√5) .■ ✓)),((x^2 −4=−4⇒x^2 =0⇒x=0 . ×)) :}$
	$x^{2} - 4 = t^{2}$ $\Rightarrow t + 2 \sqrt{t^{2} + 3} = t^{2} + 4 \Rightarrow$ $4 (t^{2} + 3) = t^{4} + 8 t^{2} + t^{2} - 2 t^{3} - 8 t + 16 \Rightarrow$ $t^{4} - 2 t^{3} + 5 t^{2} - 8 t + 4 = 0$ $\Rightarrow {(t - 1)}^{2} (t^{2} + 4) = 0 \Rightarrow t^{2} = 1, - 4$ $\Rightarrow {\begin{cases} x^{2} - 4 = 1 \Rightarrow x^{2} = 5 \Rightarrow x = \pm \sqrt{5} . ◼ ✓ \\ x^{2} - 4 = - 4 \Rightarrow x^{2} = 0 \Rightarrow x = 0 . \times \end{cases}$
	Answered by mr W last updated on 17/Jul/22

	$(\sqrt{x^{2} - 4} + 2 \sqrt{x^{2} - 1}) (2 \sqrt{x^{2} - 1} - \sqrt{x^{2} - 4}) = x^{2} (2 \sqrt{x^{2} - 1} - \sqrt{x^{2} - 4})$ $3 = 2 \sqrt{x^{2} - 1} - \sqrt{x^{2} - 4}$ $2 \sqrt{x^{2} - 1} = \sqrt{x^{2} - 4} + 3$ $x^{2} - 3 = 2 \sqrt{x^{2} - 4}$ $x^{4} - 6 x^{2} + 9 = 4 x^{2} - 16$ ${(x^{2} - 5)}^{2} = 0$ $\Rightarrow x^{2} = 5 \Rightarrow x = \pm \sqrt{5}$
	Commented by Tawa11 last updated on 17/Jul/22

	$Great sirs$
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