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Question Number 39995 by behi83417@gmail.com last updated on 14/Jul/18

Answered by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

$$\alpha +\beta =a+b+1\alpha \beta =a+b$$$$\alpha +\beta -(a+b+1)=0$$$$\alpha +\beta -\alpha \beta -1=0$$$$\alpha (1-\beta )-(1-\beta )=0$$$$(\alpha -1)(1-\beta )=0$$$$either\propto =1or\beta =1$$$$when\alpha =1\beta =a+b$$$$orwhen\beta =1\propto =a+b$$$$letconsider\alpha =1\beta =a+b$$$$given\alpha \beta .(4\alpha +9\beta )=5(\alpha +\beta )$$$$1(a+b)(4+9a+9b)=5(1+a+b)$$$$k=a+b$$$$k(4+9k)=5+5k$$$$4k+9k^2-5-5k=0$$$$9k^2-k-5=0$$$$k=\frac{1\pm \sqrt{1+180}}{18}=\frac{1\pm \sqrt{181}}{18}=a+b=\beta$$$$1)\frac{\sqrt{\alpha }+1}{\beta }+\frac{\sqrt{\beta }+1}{\alpha }$$$$=\frac{1+1}{a+b}+\frac{\sqrt{a+b}+1}{1}$$$$puttinga+b=\frac{1\pm \sqrt{181}}{18}wecanfind...thevalue..$$$$$$

Answered by MJS last updated on 15/Jul/18

$$\alpha +\beta =a+b+1$$$$\alpha \beta =a+b$$$$\alpha +\beta =\alpha \beta +1\Rightarrow \alpha =1\vee \beta =1$$$$$$$$\alpha \beta (4\alpha +9\beta )=5(\alpha +\beta )$$$$\mathrm{case}1:\alpha =1$$$${\beta }^2-\frac{1}{9}\beta -\frac{5}{9}=0$$$$\beta =\frac{1}{18}\pm \frac{\sqrt{181}}{18}$$$$\frac{\sqrt{\alpha }+1}{\beta }+\frac{\sqrt{\beta }+1}{\alpha }=$$$$=\frac{4-\sqrt{181}}{5}+\frac{\sqrt{-2+2\sqrt{181}}}{6}\mathrm{i}\vee \frac{4+\sqrt{181}}{5}+\frac{\sqrt{2+2\sqrt{181}}}{6}$$$$\alpha \sqrt{\beta }+\beta \sqrt{\alpha }+\sqrt{\alpha \beta }=$$$$=\frac{1-\sqrt{181}}{18}+\frac{\sqrt{-2+2\sqrt{181}}}{3}\mathrm{i}\vee \frac{1+\sqrt{181}}{18}+\frac{\sqrt{2+2\sqrt{181}}}{3}$$$$$$$$\mathrm{case}2:\beta =1$$$${\alpha }^2+\alpha -\frac{5}{4}=0$$$$\alpha =-\frac{1}{2}\pm \frac{\sqrt{6}}{2}$$$$\frac{\sqrt{\alpha }+1}{\beta }+\frac{\sqrt{\beta }+1}{\alpha }=$$$$=\frac{9-4\sqrt{6}}{5}+\frac{\sqrt{2+2\sqrt{6}}}{2}\mathrm{i}\vee \frac{9+4\sqrt{6}}{5}+\frac{\sqrt{-2+2\sqrt{6}}}{2}$$$$\alpha \sqrt{\beta }+\beta \sqrt{\alpha }+\sqrt{\alpha \beta }=$$$$=-\frac{1+\sqrt{6}}{2}+\mathrm{i}\sqrt{2+2\sqrt{6}}\vee \frac{-1+\sqrt{6}}{2}+\sqrt{-2+2\sqrt{6}}$$

Commented by tanmay.chaudhury50@gmail.com last updated on 15/Jul/18

$$thankusir...$$

Commented by behi83417@gmail.com last updated on 15/Jul/18

$$thankyousirMJSandsirtanmay.$$

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