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Question Number 191947 by universe last updated on 04/May/23

$$provethat$$$$\sqrt{a+b\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{...}}}}}=\frac{\sqrt{4a-3b^2}+b}{2}$$

Answered by ajfour last updated on 05/May/23

$$Lets=\sqrt{a+b\sqrt{a-bs}}=\frac{b+\sqrt{4a-3b^2}}{2}$$$$\Rightarrow s^2=a+b\sqrt{a-bs}$$$${(s^2-a)}^2=b^2(a-bs)...\left(i\right)$$$$\mathrm{\&}{(2s-b)}^2=4a-3b^2$$$$\Rightarrow 4s^2+b^2-4bs=4a-3b^2$$$$or{(s^2-a)}^2=b^2{(s-b)}^2..\left(ii\right)$$$$comparing\left(i\right)\mathrm{\&}\left(ii\right)$$$$\Rightarrow {(s-b)}^2=a-bs$$$${(s-b)}^2+b(s-b)-(a-b^2)=0$$$$s-b=\frac{-b\pm \sqrt{4a-3b^2}}{2}$$$$s=\frac{b\pm \sqrt{4a-3b^2}}{2}$$

Commented by mr W last updated on 04/May/23

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Commented by mehdee42 last updated on 05/May/23

$$dearfriend,whatisthereasonforusingequation{(2s-b)}^2=4a-3b^2?$$

Answered by universe last updated on 05/May/23

$$\sqrt{a+b\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{...}}}}}=x\left(let\right)...\left(1\right)$$$$and$$$$\sqrt{a-b\sqrt{a+b\sqrt{a-b\sqrt{a+b\sqrt{...}}}}}=y\left(let\right)...\left(2\right)$$$$by\left(1\right)$$$$\sqrt{a+by}=x\Rightarrow a+by=x^2....\left(3\right)$$$$nowby\left(2\right)$$$$\sqrt{a-bx}=y\Rightarrow a-bx=y^2....\left(4\right)$$$${eq}^n\left(3\right)-{eq}^n\left(4\right)$$$$b(x+y)=(x-y)(x+y)\Rightarrow x-y=b$$$$y=x-b$$$$by\left(3\right)$$$$a+bx-b^2=x^2\Rightarrow x^2-bx+b^2-a=0$$$$x=\frac{b+\sqrt{b^2-4(b^2-a)}}{2}$$$$x=\frac{b+\sqrt{4a-3b^2}}{2}$$

Commented by mehdee42 last updated on 05/May/23

$$abeautifulsolution$$

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