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TS EAMCET · Maths · Quadratic Equation

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

  1. A \(8 x^2+8 x+1=0\)
  2. B \(8 x^2-8 x-1=0\)
  3. C \(8 x^2-8 x+1=0\)
  4. D \(8 x^2+8 x-1=0\)
Verified Solution

Answer & Solution

Correct Answer

(C) \(8 x^2-8 x+1=0\)

Step-by-step Solution

Detailed explanation

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,…