Question

# If $\alpha ,\beta$ are the roots of the equation ${{\text{x}}^2} - 3{\text{x + 1 = 0}}$,then the equation with roots $\dfrac{1}{{(\alpha - 2)}},\dfrac{1}{{(\beta - 2)}}$ will be:-

Hint: we will proceed by writing the general expression of a quadratic equation when its two roots are given. After this, we will find the sum and product of roots of a given quadratic equation. Finally we will put these values in a simplified expression of the new quadratic equation.

Given a quadratic equation is ${{\text{x}}^2} - 3{\text{x + 1 = 0}}$ and $\alpha ,\beta$ are the roots of this equation.
The quadratic equation with roots $\dfrac{1}{{(\alpha - 2)}},\dfrac{1}{{(\beta - 2)}}$ is given by:
$({\text{x - }}\dfrac{1}{{(\alpha - 2)}})({\text{x - }}\dfrac{1}{{(\beta - 2)}}) = \dfrac{{{\text{(x(}}\alpha - 2) - 1)}}{{(\alpha - 2)}} \times \dfrac{{{\text{(x(}}\beta {\text{ - 2) - 1)}}}}{{(\beta - 2)}} = \dfrac{{{{\text{x}}^2}(\alpha \beta - 2(\alpha + \beta ) + 4 - {\text{x(}}\alpha + \beta - 4) + 1}}{{\alpha \beta - 2(\alpha + \beta ) + 4}}$ .
The above expression is in the form of $\alpha + \beta {\text{ and }}\alpha \beta$.
For quadratic equation ${{\text{x}}^2} - 3{\text{x + 1 = 0}}$:
Sum of roots = $\alpha + \beta = \dfrac{{ - {\text{b}}}}{{\text{a}}} = \dfrac{{ - ( - 3)}}{1} = 3$ .
And product of roots =$\alpha \beta = \dfrac{{\text{c}}}{{\text{a}}} = \dfrac{1}{1} = 1$
Putting the values of $\alpha + \beta {\text{ and }}\alpha \beta$ in above equation, we get:
$\dfrac{{{{\text{x}}^2}(1 - 2 \times 3 + 4) - {\text{x(3}} - 4) + 1}}{{1 - 2 \times 3 + 4}} = \dfrac{{ - {{\text{x}}^2} + {\text{x + 1}}}}{{ - 1}} = {{\text{x}}^2} - {\text{x - 1}}{\text{.}}$
Therefore the required quadratic equation is ${{\text{x}}^2} - {\text{x - 1}}{\text{.}}$
Note: In this question the important step is to convert the new quadratic equation into the form where terms like $\alpha + \beta {\text{ and }}\alpha \beta$ appear in the whole expression. You should remember the general expression of a quadratic equation which is given as :${\text{a}}{{\text{x}}^2} + {\text{bx + c = 0}}$.