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Question Number 28406 by Tinkutara last updated on 25/Jan/18

$$Imagenotgettingattached.Please$$$$seethelinkbelow.$$

Commented by Tinkutara last updated on 25/Jan/18

http://ibb.co/iaU0Jb Any method other than looking options?

Answered by Rasheed.Sindhi last updated on 26/Jan/18

$$\alpha ,\beta \mathrm{are}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2-{6\mathrm{p}}_1\mathrm{x}+2=0;$$$$\beta ,\gamma \mathrm{are}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2-{6\mathrm{p}}_2\mathrm{x}+3=0;$$$$\gamma ,\alpha \mathrm{are}\mathrm{roots}\mathrm{of}{\mathrm{x}}^2-{6\mathrm{p}}_3\mathrm{x}+6=0;$$$${\mathrm{p}}_1,{\mathrm{p}}_2\mathrm{\&}{\mathrm{p}}_3\mathrm{are}\mathrm{positive}\mathrm{then}:$$$$\mathbf{Choose}\mathbf{the}\mathbf{correct}\mathbf{answer}$$$$1.\mathrm{The}\mathrm{values}\mathrm{of}\alpha ,\beta ,\gamma \mathrm{respectively}\mathrm{are}$$$$\left(1\right)2,3,1\left(2\right)2,1,3\left(3\right)1,2,3\left(4\right)-1,-2,-3$$$$2.\mathrm{The}\mathrm{values}\mathrm{of}{\mathrm{p}}_1,{\mathrm{p}}_2,{\mathrm{p}}_3\mathrm{respectively}\mathrm{are}$$$$\left(1\right)\frac{1}{2},\frac{2}{3},\frac{5}{6}\left(2\right)1,2,5\left(3\right)6,1,4\left(4\right)2,\frac{3}{2},\frac{6}{5}$$$$-.-.-.-.-.-$$$$1.$$$$\alpha \beta =2,\beta \gamma =3,\gamma \alpha =6$$$$\mathrm{Multiplying}$$$${\alpha }^2{\beta }^2{\gamma }^2=36$$$$\alpha \beta \gamma =\pm 6$$$$\frac{\alpha \beta \gamma }{\beta \gamma }=\frac{\pm 6}{3}\Rightarrow \alpha =\pm 2$$$$\frac{\alpha \beta \gamma }{\alpha \gamma }=\frac{\pm 6}{6}\Rightarrow \beta =\pm 1$$$$\frac{\alpha \beta \gamma }{\alpha \beta }=\frac{\pm 6}{2}=\pm 3$$$$(\alpha ,\beta ,\gamma )=(2,1,3)$$$$\mathrm{Or}$$$$(\alpha ,\beta ,\gamma )=(-2,-1,-3)\left(\mathrm{Not}\mathrm{included}\mathrm{in}\mathrm{options}\right)$$$$$$$$\left(2\right)2,1,3$$$$-.-.-.-.-.-$$$$2.$$$$\alpha +\beta ={6\mathrm{p}}_1\Rightarrow {\mathrm{p}}_1=\frac{2+1}{6}=\frac{1}{2}$$$$\beta +\gamma ={6\mathrm{p}}_2\Rightarrow {\mathrm{p}}_2=\frac{1+3}{6}=\frac{2}{3}$$$$\gamma +\alpha ={6\mathrm{p}}_2\Rightarrow {\mathrm{p}}_3=\frac{3+2}{6}=\frac{5}{6}$$$$({\mathrm{p}}_1,{\mathrm{p}}_2,{\mathrm{p}}_3)=(\frac{1}{2},\frac{2}{3},\frac{5}{6})$$$$\left(1\right)\frac{1}{2},\frac{2}{3},\frac{5}{6}$$$$⋮$$

Commented by Tinkutara last updated on 26/Jan/18

Thanks Sir! Rest I solved.

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