Math Question and Answers


Question and Answers Forum

All Questions Topic List
Algebra Questions
Previous in All Question Next in All Question
Previous in Algebra Next in Algebra
Question Number 164576 by mathlove last updated on 19/Jan/22

	Answered by Rasheed.Sindhi last updated on 19/Jan/22
	$(√(2+(√3))) =a⇒(1/a)=(1/( (√(2+(√3)))))∙((√(2−(√3)))/( (√(2−(√3))))) (1/a)=(√(2−(√3))) ((√(2−(√3))))^x +((√(2+(√3))) )^x =4 ⇒a^x +(1/a^x )=4⇒a^(2x) −4a^x +1=0 a^x =((4±(√(16−4)))/2)=2±(√3) { ((2+(√3) =a^2 )),((2−(√3) =a^(−2) )) :} a^x =a^(±2) x=±2$
	$\sqrt{2 + \sqrt{3}} = a \Rightarrow \frac{1}{a} = \frac{1}{\sqrt{2 + \sqrt{3}}} \cdot \frac{\sqrt{2 - \sqrt{3}}}{\sqrt{2 - \sqrt{3}}}$ $\frac{1}{a} = \sqrt{2 - \sqrt{3}}$ ${(\sqrt{2 - \sqrt{3}})}^{x} + {(\sqrt{2 + \sqrt{3}})}^{x} = 4$ $\Rightarrow a^{x} + \frac{1}{a^{x}} = 4 \Rightarrow a^{2 x} - 4 a^{x} + 1 = 0$ $a^{x} = \frac{4 \pm \sqrt{16 - 4}}{2} = 2 \pm \sqrt{3} {\begin{cases} 2 + \sqrt{3} = a^{2} \\ 2 - \sqrt{3} = a^{- 2} \end{cases}$ $a^{x} = a^{\pm 2}$ $x = \pm 2$
Terms of Service
Privacy Policy
Contact: info@tinkutara.com

Question and Answers Forum