Thursday, 4 September 2014

polynomials - Quadratic equation including Arithmetic Progression




For a,b,c are real. Let a+b1ab,b,b+c1bc 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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