Quadratic Equations and Expressions
2.0 Relation between roots and coefficients
2.0 Relation between roots and coefficients
A root of the quadratic equation $a{x^2} + bx + c = 0$ is a number $\alpha $ (real or complex) such that $$a{\alpha ^2} + b\alpha + c = 0$$
Using discriminant $D = {b^2} - 4ac$, the roots are given by $$\begin{equation} \begin{aligned} x = \frac{{ - b \pm \sqrt D }}{{2a}} \\ {\text{ = }}\frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\\end{aligned} \end{equation} $$
If $\alpha $ and $\beta $ are the roots of the equation $$a{x^2} + bx + c = 0\quad ...(1)$$ It can also be written as $$(x - \alpha )(x - \beta ) = 0$$ or, $${x^2} - (\alpha + \beta )x + \alpha \beta = 0\quad ...(2)$$
Comparing equations $(1)$ and $(2)$,
Sum of roots i.e., $$\alpha + \beta = - \frac{b}{a} = - \frac{{{\text{coefficient of }}x}}{{{\text{coefficient of }}{x^2}}}$$ and Product of roots i.e., $$\alpha \times \beta = \frac{c}{a} = \frac{{{\text{constant term}}}}{{{\text{coefficient of }}{x^2}}}$$
Question 1. If $\alpha $ and $\beta $ are the roots of the equation $a{x^2} + bx + c = 0$, prove that $$a\left( {\frac{{{\alpha ^2}}}{\beta } + \frac{{{\beta ^2}}}{\alpha }} \right) + b\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha }} \right) = b$$
Solution: As we know that $$\alpha + \beta = - \frac{b}{a}{\text{ and }}\alpha \beta = \frac{c}{a}$$
Take L.H.S.
$$\begin{equation} \begin{aligned} a\left( {\frac{{{\alpha ^2}}}{\beta } + \frac{{{\beta ^2}}}{\alpha }} \right) + b\left( {\frac{\alpha }{\beta } + \frac{\beta }{\alpha }} \right) \\ \Rightarrow \frac{{a({\alpha ^3} + {\beta ^3}) + b({\alpha ^2} + {\beta ^2})}}{{\alpha \beta }} \\ \Rightarrow \frac{{a\left[ {{{\left( {\alpha + \beta } \right)}^3} - 3\alpha \beta (\alpha + \beta )} \right] + b\left[ {{{\left( {\alpha + \beta } \right)}^2} - 2\alpha \beta } \right]}}{{\alpha \beta }} \\ \Rightarrow \frac{{a\left[ {{{\left( { - \frac{b}{a}} \right)}^3} - 3\frac{c}{a}\left( { - \frac{b}{a}} \right)} \right] + b\left[ {{{\left( { - \frac{b}{a}} \right)}^2} - 2\frac{c}{a}} \right]}}{{\frac{c}{a}}} \\ \Rightarrow \frac{{\frac{{bc}}{a}}}{{\frac{c}{a}}} = b \\\end{aligned} \end{equation} $$
