Question-200262

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

Answered by Frix last updated on 17/Nov/23

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

Answered by AST last updated on 17/Nov/23

⇒r=((b(a+b))/(2b+a+(√(a^2 +2b^2 +2ab))))....(i) ((r(a+(√(a^2 +2b^2 +2ab))+(√2)(a+b))/2)=(((a+b)a)/2) ⇒r=(((a+b)a)/(a+(√(a^2 +2b^2 +2ab))+(√2)(a+b)))...(ii) ⇒(b/(2b+a+(√(a^2 +2b+2ab))))=(a/(a+(√(a^2 +2b^2 +2ab))+(√2)(a+b))) ⇒b(√(a^2 +2b^2 +2ab))+(√2)ab+(√2)b^2 =ab+a(√(a^2 +2b+2ab))+a^2 a=bx⇒(√((x^2 +2+2x)))+(√2)x+(√2)=x+x^2 +x(√(x^2 +2+2x)) ⇒x=−1 or x=≈1.1973682

$$\Rightarrow{r}=\frac{{b}\left({a}+{b}\right)}{\mathrm{2}{b}+{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}}}….\left({i}\right) \\ $$$$\frac{{r}\left({a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}}+\sqrt{\mathrm{2}}\left({a}+{b}\right)\right.}{\mathrm{2}}=\frac{\left({a}+{b}\right){a}}{\mathrm{2}} \\ $$$$\Rightarrow{r}=\frac{\left({a}+{b}\right){a}}{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}}+\sqrt{\mathrm{2}}\left({a}+{b}\right)}…\left({ii}\right) \\ $$$$\Rightarrow\frac{{b}}{\mathrm{2}{b}+{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{2}{ab}}}=\frac{{a}}{{a}+\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}}+\sqrt{\mathrm{2}}\left({a}+{b}\right)} \\ $$$$\Rightarrow{b}\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}^{\mathrm{2}} +\mathrm{2}{ab}}+\sqrt{\mathrm{2}}{ab}+\sqrt{\mathrm{2}}{b}^{\mathrm{2}} ={ab}+{a}\sqrt{{a}^{\mathrm{2}} +\mathrm{2}{b}+\mathrm{2}{ab}}+{a}^{\mathrm{2}} \\ $$$${a}={bx}\Rightarrow\sqrt{\left({x}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}{x}\right)}+\sqrt{\mathrm{2}}{x}+\sqrt{\mathrm{2}}={x}+{x}^{\mathrm{2}} +{x}\sqrt{{x}^{\mathrm{2}} +\mathrm{2}+\mathrm{2}{x}} \\ $$$$\Rightarrow{x}=−\mathrm{1}\:{or}\:{x}=\approx\mathrm{1}.\mathrm{1973682} \\ $$

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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