Question

# If $\alpha$ and $\beta$ are the roots of the quadratic equation ${{x}^{2}}-x-1=0$ , then the quadratic equation whose roots are $\dfrac{1+\alpha }{2-\alpha },\dfrac{1+\beta }{2-\beta }$ is A.${{z}^{2}}+z+1=0$ B. ${{z}^{2}}-7z+1=0$              C. ${{z}^{2}}+7z+1=0$             D. ${{z}^{2}}+7z-1=0$

Hint: - For solving this particular type of question, we need to follow the following procedure to get to the solution which is as follows
1.First we will find the roots of the given quadratic equation.
2.Then we will let one of the roots of the given equation to be ${{\alpha }_{o}}$ and then we will let one of the new roots be ${{\alpha }_{n}}$ .
3.As we know that ${{\alpha }_{n}}=\dfrac{1 + \alpha }{2-\alpha }$ and ${{\alpha }_{o}}=\alpha$ as given in the question.
4.Now, we will put the value of ${{\alpha }_{o}}$ in ${{\alpha }_{n}}$ .
5.Now, we will solve for ${{\alpha }_{o}}$ and get it in terms of ${{\alpha }_{n}}$ and then put this value in the given quadratic equation.
6.On simplifying, we will get the new quadratic equation of which ${{\alpha }_{n}}$ is a root.

As mentioned in the question, we have to find the quadratic equation of which ${{\alpha }_{n}}$ is a root.
${{\alpha }_{n}}=\dfrac{1+{{\alpha }_{o}}}{2-{{\alpha }_{o}}}$
${{\alpha }_{o}}=\dfrac{2{{\alpha }_{n}}-1}{{{\alpha }_{n}}+1}$
Now, as we have got ${{\alpha }_{o}}$ in terms of ${{\alpha }_{n}}$ , now, we will put this value in the given quadratic equation to get the new quadratic equation which is as follows
\begin{align} & \alpha {{{}_{o}}^{2}}-{{\alpha }_{0}}-1=0 \\ & {{\left( \dfrac{2{{\alpha }_{n}}-1}{{{\alpha }_{n}}+1} \right)}^{2}}-\left( \dfrac{2{{\alpha }_{n}}-1}{{{\alpha }_{n}}+1} \right)-1=0 \\ & {{\left( 2{{\alpha }_{n}}-1 \right)}^{2}}-\left( 2{{\alpha }_{n}}-1 \right)\left( {{\alpha }_{n}}+1 \right)-{{\left( {{\alpha }_{n}}+1 \right)}^{2}}=0 \\ & \alpha {{{}_{n}}^{2}}-7{{\alpha }_{n}}+1=0 \\ \end{align}
${{z}^{2}}-7z+1=0$.