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
![α+β=((−b)/a) αβ=(c/a) ((α+β)/(αβ))=((−b)/c) α+β=((−b)/a) α^2 +β^2 =(b^2 /a^2 )−((2c)/a)=((b^2 −2ac)/a^2 ) given 2b=a+c [since a,b and c are inA.P] (((−b)/a))^2 =((−b)/c)×((b^2 −2ac)/a^2 ) ((a^2 +c^2 +2ac)/(4a^2 ))=((−a−c)/(2c))×((((a^2 +c^2 +2ac)/4)−2ac)/a^2 ) ((a^2 +c^2 +2ac)/4)=((−a−c)/(2c))×((a^2 +c^2 −2ac)/4) c^2 ((a^2 /c^2 )+1+2(a/c))=((−1)/2)((a/c)+1)×(c^2 /4)((a^2 /c^2 )+1−((2a)/c)) (a/c)=k k^2 +1+2k=((−1)/8)(k+1)(k^2 +1−2k) 8k^2 +8+16k=−1(k^3 +k−2k^2 +k^2 +1−2k) 8k^2 +8+16k=−k^3 +k^2 +k−1 k^3 +7k^2 +15k+9=0 k^3 +k^2 +6k^2 +6k+9k+9=0 k^2 (k+1)+6k(k+1)+9(k+1)=0 (k+1)(k^2 +6k+9)=0 (k+1)(k+3)(k+3)=0 so k=−1 or k=−3 (a/c)=−1 or (a/c)=−3](https://www.tinkutara.com/question/Q49693.png)
$$\alpha+\beta=\frac{−{b}}{{a}}\:\:\:\alpha\beta=\frac{{c}}{{a}} \\ $$$$\frac{\alpha+\beta}{\alpha\beta}=\frac{−{b}}{{c}}\:\:\:\:\alpha+\beta=\frac{−{b}}{{a}}\:\:\:\alpha^{\mathrm{2}} +\beta^{\mathrm{2}} =\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\mathrm{2}{c}}{{a}}=\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{a}^{\mathrm{2}} } \\ $$$${given}\:\:\mathrm{2}{b}={a}+{c}\:\left[{since}\:{a},{b}\:{and}\:{c}\:{are}\:{inA}.{P}\right] \\ $$$$\left(\frac{−{b}}{{a}}\right)^{\mathrm{2}} =\frac{−{b}}{{c}}×\frac{{b}^{\mathrm{2}} −\mathrm{2}{ac}}{{a}^{\mathrm{2}} } \\ $$$$\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}{\mathrm{4}{a}^{\mathrm{2}} }=\frac{−{a}−{c}}{\mathrm{2}{c}}×\frac{\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}{\mathrm{4}}−\mathrm{2}{ac}}{{a}^{\mathrm{2}} } \\ $$$$\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} +\mathrm{2}{ac}}{\mathrm{4}}=\frac{−{a}−{c}}{\mathrm{2}{c}}×\frac{{a}^{\mathrm{2}} +{c}^{\mathrm{2}} −\mathrm{2}{ac}}{\mathrm{4}} \\ $$$${c}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\mathrm{1}+\mathrm{2}\frac{{a}}{{c}}\right)=\frac{−\mathrm{1}}{\mathrm{2}}\left(\frac{{a}}{{c}}+\mathrm{1}\right)×\frac{{c}^{\mathrm{2}} }{\mathrm{4}}\left(\frac{{a}^{\mathrm{2}} }{{c}^{\mathrm{2}} }+\mathrm{1}−\frac{\mathrm{2}{a}}{{c}}\right) \\ $$$$\frac{{a}}{{c}}={k} \\ $$$${k}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}{k}=\frac{−\mathrm{1}}{\mathrm{8}}\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\mathrm{8}{k}^{\mathrm{2}} +\mathrm{8}+\mathrm{16}{k}=−\mathrm{1}\left({k}^{\mathrm{3}} +{k}−\mathrm{2}{k}^{\mathrm{2}} +{k}^{\mathrm{2}} +\mathrm{1}−\mathrm{2}{k}\right) \\ $$$$\mathrm{8}{k}^{\mathrm{2}} +\mathrm{8}+\mathrm{16}{k}=−{k}^{\mathrm{3}} +{k}^{\mathrm{2}} +{k}−\mathrm{1} \\ $$$${k}^{\mathrm{3}} +\mathrm{7}{k}^{\mathrm{2}} +\mathrm{15}{k}+\mathrm{9}=\mathrm{0} \\ $$$${k}^{\mathrm{3}} +{k}^{\mathrm{2}} +\mathrm{6}{k}^{\mathrm{2}} +\mathrm{6}{k}+\mathrm{9}{k}+\mathrm{9}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} \left({k}+\mathrm{1}\right)+\mathrm{6}{k}\left({k}+\mathrm{1}\right)+\mathrm{9}\left({k}+\mathrm{1}\right)=\mathrm{0} \\ $$$$\left({k}+\mathrm{1}\right)\left({k}^{\mathrm{2}} +\mathrm{6}{k}+\mathrm{9}\right)=\mathrm{0} \\ $$$$\left({k}+\mathrm{1}\right)\left({k}+\mathrm{3}\right)\left({k}+\mathrm{3}\right)=\mathrm{0} \\ $$$${so}\:{k}=−\mathrm{1}\:\:\:{or}\:{k}=−\mathrm{3} \\ $$$$\frac{{a}}{{c}}=−\mathrm{1}\:\:\:{or}\:\frac{{a}}{{c}}=−\mathrm{3}\:\: \\ $$$$ \\ $$$$ \\ $$
Commented by Pk1167156@gmail.com last updated on 10/Dec/18
Thank you very much sir.