Question

If $\alpha ,\beta $ are the roots of the equation $a{x^2} + bx + c = 0$ , then $\dfrac{\alpha }{{a\beta + b}} + \dfrac{\beta }{{a\alpha + b}}$ is equal to
A) $\dfrac{2}{a}$
B) $\dfrac{2}{b}$
C) $\dfrac{2}{c}$
D) $\dfrac{{ - 2}}{a}$

Answer 1

Hint: Here in this question, we have to justify the above equation as for which simply we have to find simply sum of roots and the product of the roots from the use of $\alpha,\beta $ out of which we simply put the value in the equation we get the solution of the equation.

Formula Used: Sum of roots $(\alpha + \beta ) = \dfrac{{ - b}}{a}$
Product of roots $(\alpha \beta ) = \dfrac{c}{a}$
${\alpha ^2} + {\beta ^2} = \dfrac{{{b^2} - 2ac}}{{{a^2}}}$

Complete step by step Solution:
As we have to find the solution to the equation,
$\dfrac{\alpha }{{a\beta + b}} + \dfrac{\beta }{{a\alpha + b}}$
For which, we just take L.C.M to make them close to capturing the calculation,
$\dfrac{\alpha }{{a\beta + b}} + \dfrac{\beta }{{a\alpha + b}} = \dfrac{{\alpha (a\alpha + b) + \beta (a\beta + b)}}{{(a\beta + b)(a\alpha + b)}}$
By doing calculations as perusing the formulas which are as mentioned in the formula used section we get the result of the equation as,
$ = \dfrac{{{b^2} - ac - {b^2}}}{{{a^2}c - a{b^2} + a{b^2}}}$
By doing a further calculation of the above equation, we get the solution of the equation as,
$ = \dfrac{{ - 2ac}}{{{a^2}c}}$
By canceling out the common factors from the numerator and denominator we get the result as,
$ = - \dfrac{2}{a}$
From the whole conclusion, we get the answer as $ - \dfrac{2}{a}$ .
Therefore, the correct answer of the equation $\dfrac{\alpha }{{a\beta + b}} + \dfrac{\beta }{{a\alpha + b}}$ is $ - \dfrac{2}{a}$ .

Therefore, the correct option is (D).

Note:Some important points about the roots of the equation which we must know about it. The physical importance of the roots is that the graph of an equation intersects the x-axis at the roots. In the Cartesian plane, the real line is represented by the x-axis. As a result, the equation cannot be factorized if it has unreal roots because it won't cross the x-axis.