Question

Let $\alpha \neq \beta$ satisfy $\alpha^{2}+1=6 \alpha, \beta^{2}+1=6 \beta$.
Then, the quadratic equation whose roots are $\frac{\alpha}{\alpha+1}, \frac{\beta}{\beta+1}$, is

• A. $8 x^{2}+8 x+1=0$
• B. $8 x^{2}-8 x-1=0$
• C. $8 x^{2}-8 x+1=0$
• D. $8 x^{2}+8 x-1=0$

Answer 1

Answer: (C)

We have,
$\alpha^{2}+1=6 \,\alpha$ and $\beta^{2}+1=6\,\beta$
Since, $\alpha, \beta$ are the roots of the equation
$x^{2}-6 x+1=0$
$\therefore x=\frac{\alpha}{\alpha+1} $
$\Rightarrow \alpha=\frac{X}{1-X}$
Hence, required quadratic equation whose roots
are $\frac{\alpha}{\alpha+1}, \frac{\beta}{\beta+1}$, is $\left(\frac{x}{1-x}\right)^{2}-6\left(\frac{x}{1-x}\right)+1=0$
$\Rightarrow x^{2}-6 x(1-x)+1(1-x)^{2}=0$
$\Rightarrow x^{2}-6 x+6 x^{2}+1+x^{2}-2 x=0$
$\Rightarrow 8 x^{2}-8 x+1=0$