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Question Number 165830 by HongKing last updated on 09/Feb/22

$$\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}-\sqrt{\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}}=0$$$$\left(\mathrm{a}\right)9\left(\mathrm{b}\right)3\left(\mathrm{c}\right)6\left(\mathrm{d}\right)1\left(\mathrm{e}\right)12$$

Answered by Rasheed.Sindhi last updated on 09/Feb/22

$$\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}-\sqrt{\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}}=0$$$$\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}=\sqrt{\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}}=\mathrm{b}\left(\mathrm{say}\right)$$$$8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}=\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}$$$$8\mathrm{a}+\mathrm{b}=\mathrm{ab}$$$$8\mathrm{a}=\mathrm{b}(\mathrm{a}-1)$$$$\mathrm{b}=\frac{8\mathrm{a}}{\mathrm{a}-1}............\left(\mathrm{i}\right)$$$$\mathrm{Again},$$$$\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}=\sqrt{\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}}$$$$8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}=\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}$$$$\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}=\mathrm{ab}-8\mathrm{a}$$$$8\mathrm{a}+\sqrt{8\mathrm{a}+...}={(\mathrm{ab}-8\mathrm{a})}^2$$$$\sqrt{8\mathrm{a}+...}={(\mathrm{ab}-8\mathrm{a})}^2-8\mathrm{a}$$$$\mathrm{b}={(\mathrm{ab}-8\mathrm{a})}^2-8\mathrm{a}$$$$\frac{8\mathrm{a}}{\mathrm{a}-1}={(\mathrm{a}\left(\frac{8\mathrm{a}}{\mathrm{a}-1}\right)-8\mathrm{a})}^2-8\mathrm{a}[\because \mathrm{b}=\frac{8\mathrm{a}}{\mathrm{a}-1}\mathrm{from}\left(\mathrm{i}\right)]$$$$\frac{8\mathrm{a}}{\mathrm{a}-1}={\left(\frac{{8\mathrm{a}}^2-{8\mathrm{a}}^2+8\mathrm{a}}{\mathrm{a}-1}\right)}^2-8\mathrm{a}$$$$\frac{8\mathrm{a}}{\mathrm{a}-1}={\left(\frac{8\mathrm{a}}{\mathrm{a}-1}\right)}^2-8\mathrm{a}$$$$8\mathrm{a}(\mathrm{a}-1)={64\mathrm{a}}^2-8\mathrm{a}{(\mathrm{a}-1)}^2$$$${64\mathrm{a}}^2=8\mathrm{a}(\mathrm{a}-1)+8\mathrm{a}{(\mathrm{a}-1)}^2$$$$=8\mathrm{a}(\mathrm{a}-1)(1+\mathrm{a}-1)$$$$={8\mathrm{a}}^2(\mathrm{a}-1)$$$$\mathrm{a}-1=8$$$$\mathrm{a}=9$$

Commented by Tawa11 last updated on 10/Feb/22

$$\mathrm{Weldone}\mathrm{sir}$$

Commented by Rasheed.Sindhi last updated on 10/Feb/22

$$\mathbb{T}\mathrm{han}\mathbb{k}\mathrm{s}\mathrm{miss}!$$

Answered by naka3546 last updated on 09/Feb/22

$$\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}-\sqrt{\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}}=0$$$$\Rightarrow \sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}=\sqrt{\mathrm{a}\sqrt{\mathrm{a}\sqrt{\mathrm{a}...}}}$$$$\Rightarrow \sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+\sqrt{8\mathrm{a}+...}}}=\mathrm{a}$$$$\Rightarrow \sqrt{8\mathrm{a}+\mathrm{a}}=\mathrm{a}$$$$\Rightarrow \sqrt{9\mathrm{a}}=\mathrm{a}$$$$\Rightarrow 9\mathrm{a}={\mathrm{a}}^2(\mathrm{a}>0)$$$$\Rightarrow \mathrm{a}=9$$

Commented by Tawa11 last updated on 10/Feb/22

$$\mathrm{Weldone}\mathrm{sir}$$

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