Q77 / 160 · TS EAMCET 2015 · Mathematics

Let $\alpha \neq \beta \quad$ 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
    $8 x^2-8 x-1=0$
  • Correct answer
    C
    $8 x^2-8 x+1=0$
  • D
    $8 x^2+8 x-1=0$

Solution

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

Hence, the correct answer is option: 'C'.

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