Question

For $a, b, c \in R-\{0\}$, let $\frac{a+b}{1-a b}, b, \frac{b+c}{1-b c}$ are in A.P. If $\alpha, \beta$ are the roots of the quadratic equation $2 a c x^2+2 a b c x+(a+c)=0$, then find the value of $(1+\alpha)(1+\beta)$.

Answer 1

Given $\frac{a+b}{1-a b}, b, \frac{b+c}{1-b c}$ are in A.P.
$\Rightarrow b -\frac{ a + b }{1- ab }=\frac{ b + c }{1- bc }- b \Rightarrow \frac{- a \left( b ^2+1\right)}{1- ab }=\frac{ c \left( b ^2+1\right)}{1- bc } \Rightarrow a + c =2 abc$
Now, given quadratic equation is
$2 a c x^2+2 a b c x+2 a b c=0 $ (Substituting $a+c=2 a b c$ and then cancelling 2ac)

As $\alpha+\beta=-b, \alpha \beta=b$
$\therefore (1+\alpha)(1+\beta)=(\alpha+\beta)+(\alpha \beta)+1=-b+b+1=1 $