Question

If $\alpha $ and $\beta $ are the roots of $a{x^2} + bx + c = 0$, find the values of the following. $\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}$

Answer 1

Hint: Before approaching this question prior knowledge of quadratic equation is must, put the values of $\alpha $ and $\beta $ in the quadratic equation and make the equation, use this information to approach towards the solution to the problem

Complete step-by-step solution -
According to the given information we have quadratic equation $a{x^2} + bx + c = 0$ which have roots $\alpha $ and $\beta $
Substituting the value of x = $\alpha $ in the given quadratic equation i.e. $a{x^2} + bx + c = 0$ we get
$a{\left( \alpha \right)^2} + b\left( \alpha \right) + c = 0$
$ \Rightarrow $\[a{\alpha ^2} + b\alpha = - c\] …………….(eq.1)
Now for x = $\beta $
\[a{\left( \beta \right)^2} + b\left( \beta \right) + c = 0\]
$ \Rightarrow $\[a{\beta ^2} + b\beta = - c\] …………..(eq.2)
As we know that both roots of a quadratic equation so for the roots of quadratic equation $\alpha + \beta = \dfrac{{ - b}}{a}$ and $\alpha \beta = \dfrac{c}{a}$
Now finding the value of $\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}$
Now substituting the given values in the above equation
$\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{\beta \left( {a\beta + b} \right) + \alpha \left( {a\alpha + b} \right)}}{{\left( {a\beta + b} \right)\left( {a\alpha + b} \right)}}$
$ \Rightarrow $$\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{a{\beta ^2} + b\beta + a{\alpha ^2} + b\alpha }}{{{a^2}\alpha \beta + ab\alpha + ab\beta + {b^2}}}$ …………(eq. 3)
Substituting the given values and values form equation 1 and equation 2 to equation 3
$ \Rightarrow $$\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - c - c}}{{{a^2}\left( {\dfrac{c}{a}} \right) + ab\left( {\dfrac{{ - b}}{a}} \right) + {b^2}}}$
$ \Rightarrow $$\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - 2c}}{{ac - {b^2} + {b^2}}}$
$ \Rightarrow $$\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - 2c}}{{ac}}$
$ \Rightarrow $$\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}} = \dfrac{{ - 2}}{a}$
Hence the value of $\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}$ is $\dfrac{{ - 2}}{a}$.

Note: In the above question which was based on quadratic equation so let’s discuss about the quadratic equation which can be explained as the equation which consist of at least one squared term also this equation is called as second degree equation. The general representation of quadratic equation is $a{x^2} + bx + c = 0$ where a, b and c are the constant of quadratic equation and x represents the unknown variable.