For a,b,c are real. Let a+b1−ab,b,b+c1−bc be in arithmetic progression . If α,β are roots of equation 2acx2+2abcx+(a+c)=0 then find the value of (1+α)(1+β)
Answer
The A.P. condition gives 2abc=a+c on simplifying, so the quadratic is effectively x2+bx+b=0. As (1+α)(1+β)=1+(α+β)+(αβ), using Vieta we have this equal to 1+(−b)+b=1.
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