Question

Let $f (x) = ax^2 + bx + c, a \ne 0$ and $\Delta =b^2 -4ac.$ If $a+ \beta,$ $a^2+ \beta^2$ and $a^3+ \beta^3$ are in GP, then

• A. $\Delta = 0$
• B. $b\Delta = 0$
• C. $c\Delta = 0$
• D. $bc \ne 0$

Answer 1

Answer: (C)

$c\Delta = 0$

Answer 2

Since, $(a+\beta) , (a^2+\beta^2), (a^3+\beta^3)$ are in GP
\Rightarrow \hspace20mm (a^2+\beta^2)^2 =(a+\beta) (a^3+\beta^3)
$\Rightarrow \, \, \, \, \, \, \, a^4+\beta^4+2a^2 \beta^2 = a^4+ \beta^4 + a\beta^3+\beta a^3$
$\Rightarrow \, \, \, \, a\beta (a^2 +\beta^2 - 2a\beta) = 0$
\Rightarrow \hspace15mm \, \, \, a\beta (a-\beta)^2 = 0
\Rightarrow \hspace30mm \, \, a\beta = 0 \, \, \, \, \, or \, \, \, \, \, a = \beta
\Rightarrow \hspace33mm \frac{c}{a} = 0 \, \, \, \, \, or \, \, \, \, \, \Delta = 0
\Rightarrow \hspace33mm c \Delta = 0 \, \, \, \, \,