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Question Number 200262 by sonukgindia last updated on 16/Nov/23

Answered by Frix last updated on 17/Nov/23

$$\mathrm{I}\mathrm{get}\frac{a}{b}=-1+\sqrt{2+2\sqrt{2}}\approx 1.19737$$

Answered by mr W last updated on 17/Nov/23

$$\frac{a+\sqrt{{(a+b)}^2+b^2}+\sqrt{2}(a+b)}{a}=\frac{b+\sqrt{{(a+b)}^2+b^2}+(a+b)}{b}$$$$let\lambda =\frac{a}{b}$$$${\lambda }^2-(\sqrt{2}-1)\lambda -\sqrt{2}=(1-\lambda )\sqrt{{\lambda }^2+2\lambda +2}$$$$(\sqrt{2}-1){\lambda }^2+2(\sqrt{2}-1)\lambda -(3-\sqrt{2})=0$$$$\Rightarrow \lambda =\frac{a}{b}=\sqrt{2(\sqrt{2}+1)}-1✓$$

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