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

Answered by Frix last updated on 17/Nov/23

I get (a/b)=−1+(√(2+2(√2)))≈1.19737

$$\mathrm{I}\:\mathrm{get}\:\frac{{a}}{{b}}=−\mathrm{1}+\sqrt{\mathrm{2}+\mathrm{2}\sqrt{\mathrm{2}}}\approx\mathrm{1}.\mathrm{19737} \\ $$

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

((a+(√((a+b)^2 +b^2 ))+(√2)(a+b))/a)=((b+(√((a+b)^2 +b^2 ))+(a+b))/b) let λ=(a/b) λ^2 −((√2)−1)λ−(√2)=(1−λ)(√(λ^2 +2λ+2)) ((√2)−1)λ^2 +2((√2)−1)λ−(3−(√2))=0 ⇒λ=(a/b)=(√(2((√2)+1)))−1 ✓

$$\frac{{a}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{\mathrm{2}}\left({a}+{b}\right)}{{a}}=\frac{{b}+\sqrt{\left({a}+{b}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} }+\left({a}+{b}\right)}{{b}} \\ $$$${let}\:\lambda=\frac{{a}}{{b}} \\ $$$$\lambda^{\mathrm{2}} −\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\lambda−\sqrt{\mathrm{2}}=\left(\mathrm{1}−\lambda\right)\sqrt{\lambda^{\mathrm{2}} +\mathrm{2}\lambda+\mathrm{2}} \\ $$$$\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\lambda^{\mathrm{2}} +\mathrm{2}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)\lambda−\left(\mathrm{3}−\sqrt{\mathrm{2}}\right)=\mathrm{0} \\ $$$$\Rightarrow\lambda=\frac{{a}}{{b}}=\sqrt{\mathrm{2}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)}−\mathrm{1}\:\checkmark \\ $$

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