(2+(√3))^x^2 + (2−(√3))^x^2 = 4


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Question Number 103914 by bobhans last updated on 18/Jul/20

${(2 + \sqrt{3})}^{x^{2}} + {(2 - \sqrt{3})}^{x^{2}} = 4$
Commented by som(math1967) last updated on 18/Jul/20

$let {(2 + \sqrt{3})}^{x^{2}} = a$ $∴ {(2 - \sqrt{3})}^{x^{2}} = \frac{1}{a}$ $\Rightarrow a^{2} - 4 a + 1 = 0$ $\Rightarrow a = (2 \pm \sqrt{3})$ ${(2 + \sqrt{3})}^{x^{2}} = (2 + \sqrt{3})$ $\Rightarrow x = \pm 1$ ${(2 + \sqrt{3})}^{x^{2}} = {(2 + \sqrt{3})}^{- 1}$ $x = i$
Commented by bemath last updated on 18/Jul/20

$t h a n k y o u b o t h$
	Answered by OlafThorendsen last updated on 18/Jul/20
	$X = (2+(√3))^x^2 (2−(√3))^x^2 = ((1/(2+(√3))))^x^2 = (1/((2+(√3))^x^2 )) = (1/X) then X+(1/X) = 4 X^2 −4X+1 = 0 Δ = (−4)^2 −4(1)(1) = 12 X = ((−(−4)±(√(12)))/(2(1))) = 2±(√3) 1) X = 2+(√3) x^2 = 1, x=±1 2) X = 2−(√3) 2−(√3) = (2+(√3))^x^2 (1/(2+(√3))) = (2+(√3))^x^2 (2+(√3))^(x^2 +1) = 1 x^2 +1 = 0 x = ±i S = {±1;±i}$
	$X = {(2 + \sqrt{3})}^{x^{2}}$ ${(2 - \sqrt{3})}^{x^{2}} = {(\frac{1}{2 + \sqrt{3}})}^{x^{2}} = \frac{1}{{(2 + \sqrt{3})}^{x^{2}}} = \frac{1}{X}$ $then X + \frac{1}{X} = 4$ $X^{2} - 4 X + 1 = 0$ $Δ = {(- 4)}^{2} - 4 (1) (1) = 12$ $X = \frac{- (- 4) \pm \sqrt{12}}{2 (1)} = 2 \pm \sqrt{3}$ $1) X = 2 + \sqrt{3}$ $x^{2} = 1, x = \pm 1$ $2) X = 2 - \sqrt{3}$ $2 - \sqrt{3} = {(2 + \sqrt{3})}^{x^{2}}$ $\frac{1}{2 + \sqrt{3}} = {(2 + \sqrt{3})}^{x^{2}}$ ${(2 + \sqrt{3})}^{x^{2} + 1} = 1$ $x^{2} + 1 = 0$ $x = \pm i$ $S = {\pm 1; \pm i}$
	Answered by Rauny last updated on 18/Jul/20

	${(2 + \sqrt{3})}^{x^{2}} + {(2 - \sqrt{3})}^{x^{2}} = 4$ $t := 2 + \sqrt{3}$ $\frac{1}{2 + \sqrt{3}} = \frac{2 - \sqrt{3}}{(2 + \sqrt{3}) (2 - \sqrt{3})} = 2 - \sqrt{3}$ $∴ 2 - \sqrt{3} = \frac{1}{t}$ $t^{x^{2}} + t^{- x^{2}} - 4 = 0$ $u := t^{x^{2}} (\forall u > 0)$ $u - 4 + \frac{1}{u} = 0$ $u^{2} - 4 u + 1 = 0$ $u = 2 \pm \sqrt{3}$ $t^{x^{2}} = 2 \pm \sqrt{3}$ $x^{2} = \log_{2 + \sqrt{3}} (2 \pm \sqrt{3})$ $if u = 2 + \sqrt{3},$ $x^{2} = 1$ $x = \pm 1$ $if u = 2 - \sqrt{3},$ $x^{2} = \log_{2 + \sqrt{3}} \frac{1}{2 + \sqrt{3}} = - 1 (∵ 2 - \sqrt{3} = \frac{1}{2 + \sqrt{3}})$ $x = \pm i$ $∴ x = \pm 1 or \pm i ◼$
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