Question Number 49637 by Pk1167156@gmail.com last updated on 08/Dec/18
Answered by tanmay.chaudhury50@gmail.com last updated on 09/Dec/18
$$\alpha+\beta=\frac{−{b}}{{a}}\:\:\:\alpha\beta=\frac{{c}}{{a}} \\ $$$$\frac{\alpha+\beta}{\alpha\beta}=\frac{−{b}}{{c}}\:\:\:\:\alpha+\beta=\frac{−{b}}{{a}}\:\:\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}=\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{a}^{\mathrm{2}} } \\ $$$${given}\:\:\mathrm{2}{b}={a}+{c}\:\left[{since}\:{a},{b}\:{and}\:{c}\:{are}\:{inA}.{P}\right] \\ $$$$\left(\frac{−{b}}{{a}}\right)^{\mathrm{2}} =\frac{−{b}}{{c}}×\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{a}^{\mathrm{2}} } \\ $$$$\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}{\mathrm{4}{a}^{\mathrm{2}} }=\frac{−{a}−{c}}{\mathrm{2}{c}}×\frac{\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}{\mathrm{4}}−\mathrm{2}{ac}}{{a}^{\mathrm{2}} } \\ $$$$\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}{\mathrm{4}}=\frac{−{a}−{c}}{\mathrm{2}{c}}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ac}}{\mathrm{4}} \\ $$$${c}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\mathrm{1}+\mathrm{2}\frac{{a}}{{c}}\right)=\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{{a}}{{c}}+\mathrm{1}\right)×\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{2}{a}}{{c}}\right) \\ $$$$\frac{{a}}{{c}}={k} \\ $$$${k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{k}=\frac{−\mathrm{1}}{\mathrm{8}}\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\mathrm{8}{k}^{\mathrm{2}} +\mathrm{8}+\mathrm{16}{k}=−\mathrm{1}\left({k}^{\mathrm{3}} +{k}−\mathrm{2}{k}^{\mathrm{2}} +{k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\mathrm{8}{k}^{\mathrm{2}} +\mathrm{8}+\mathrm{16}{k}=−{k}^{\mathrm{3}} +{k}^{\mathrm{2}} +{k}−\mathrm{1} \\ $$$${k}^{\mathrm{3}} +\mathrm{7}{k}^{\mathrm{2}} +\mathrm{15}{k}+\mathrm{9}=\mathrm{0} \\ $$$${k}^{\mathrm{3}} +{k}^{\mathrm{2}} +\mathrm{6}{k}^{\mathrm{2}} +\mathrm{6}{k}+\mathrm{9}{k}+\mathrm{9}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)+\mathrm{6}{k}\left({k}+\mathrm{1}\right)+\mathrm{9}\left({k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} +\mathrm{6}{k}+\mathrm{9}\right)=\mathrm{0} \\ $$$$\left({k}+\mathrm{1}\right)\left({k}+\mathrm{3}\right)\left({k}+\mathrm{3}\right)=\mathrm{0} \\ $$$${so}\:{k}=−\mathrm{1}\:\:\:{or}\:{k}=−\mathrm{3} \\ $$$$\frac{{a}}{{c}}=−\mathrm{1}\:\:\:{or}\:\frac{{a}}{{c}}=−\mathrm{3}\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by Pk1167156@gmail.com last updated on 10/Dec/18
Thank you very much sir.