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For $a, b, c$ are real. Let $\frac{a+b}{1-ab}, b, \frac{b+c}{1-bc}$ be in arithmetic progression . If $\alpha, \beta$ are roots of equation $2acx^2+2abcx+(a+c) =0$ then find the value of $(1+\alpha)(1+\beta)$

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    $\begingroup$ What have you tried? $\endgroup$ – Yuta May 2 at 13:30
  • $\begingroup$ What means "are in ap."? $\endgroup$ – Dietrich Burde May 2 at 13:31
  • $\begingroup$ Yes i have tried that if abc are in ap then 2b=a+c. I tried to get a relation between a b and c. But i got to nowhere $\endgroup$ – Suman Chandra May 2 at 13:31
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    $\begingroup$ ap is arithmetic progression $\endgroup$ – Suman Chandra May 2 at 13:32
  • $\begingroup$ @Suman Chandra So $\frac{a+b}{1-ab}$, $b$ and $\frac{b+c}{1-bc}$ are in ap gives $\frac{a+b}{1-ab}+\frac{b+c}{1-bc}=2b$. This condition may be useful. $\endgroup$ – Yuta May 2 at 13:34
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The A.P. condition gives $2abc=a+c$ on simplifying, so the quadratic is effectively $x^2+bx+b=0$. As $(1+\alpha)(1+\beta) = 1+(\alpha+\beta) + (\alpha \beta)$, using Vieta we have this equal to $1+(-b)+b=1$.

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Hint: $$(a+b)(1-ac)+(b+c)(1-ab)=2b(1-ab)(1-ac)$$ gives us $$-(1+b^2)(-a-c+2abc)=0$$,so it must be $$a+c=2abc$$ Now we must solve the quadratic $$x_{1,2}=-\frac{b}{2}\pm\sqrt{\frac{b^2}{4}-\frac{a+c}{2ac}}$$ Plugging the term $$a+c=2abc$$ into the solution of the quadric equation we get $$x_{1,2}=-\frac{b}{2}\pm\sqrt{\frac{b^2}{4}-b}$$

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