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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

α+β=−baαβ=caα+βαβ=−bcα+β=−baα2+β2=b2a2−2ca=b2−2aca2given2b=a+c[sincea,bandcareinA.P](−ba)2=−bc×b2−2aca2a2+c2+2ac4a2=−a−c2c×a2+c2+2ac4−2aca2a2+c2+2ac4=−a−c2c×a2+c2−2ac4c2(a2c2+1+2ac)=−12(ac+1)×c24(a2c2+1−2ac)ac=kk2+1+2k=−18(k+1)(k2+1−2k)8k2+8+16k=−1(k3+k−2k2+k2+1−2k)8k2+8+16k=−k3+k2+k−1k3+7k2+15k+9=0k3+k2+6k2+6k+9k+9=0k2(k+1)+6k(k+1)+9(k+1)=0(k+1)(k2+6k+9)=0(k+1)(k+3)(k+3)=0sok=−1ork=−3ac=−1orac=−3

Commented by Pk1167156@gmail.com last updated on 10/Dec/18

Thank you very much sir.

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