Question

If \[\alpha \] and $\beta $ are the zeros of the quadratic polynomial $f(x) = a{x^2} + bx + c$ , then evaluate: \[\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}\]

Answer 1

Hint: First of all, we will find out the sum and products of roots for the given equation, and simplify the terms we want to evaluate, and put the values of sum and products of roots in the simplified equation, then we will get the final value.

Complete step-by-step answer:
As per the question, \[\alpha \] and $\beta $ are the zeros or roots of the quadratic polynomial $f(x) = a{x^2} + bx + c$, this is a general quadratic equation so, their roots are also general.
Therefore, as per the given equation sum and product of these roots will be as below:
As we know the sum of the root of an equation, $\alpha + \beta = \dfrac{{ - b}}{a}$
And product of roots, $\alpha \beta = \dfrac{c}{a}$
Now, we will simplify the equation which we want to evaluate by taking L.C.M.
\[\dfrac{\beta }{{a\alpha + b}} + \dfrac{\alpha }{{a\beta + b}}\]
While taking L.C.M. In numerator, we will multiply $\beta $ with $(a\beta + b)$ , $\alpha $with $(a\alpha + b)$ and in denominator we will multiply $(a\alpha + b)$ with $(a\beta + b)$
Here, the result of L.C.M.
$ = \dfrac{{a{\beta ^2} + \beta b + a{\alpha ^2} + b\alpha }}{{{a^2}\alpha \beta + a\alpha b + ab\beta + {b^2}}}$
Now, we will take commons from the numerator: $a$ common from first, third terms and take $b$ common from second, fourth terms.
Take commons from the denominator: $ab$ second and third terms.
After that we get $ = \dfrac{{a({\beta ^2} + {\alpha ^2}) + b(\alpha + \beta )}}{{ab(\alpha + \beta ) + {a^2}\alpha \beta + {b^2}}}$
We have an identity ${a^2} + {b^2} = {(a + b)^2} - 2ab$, which we can put in the first term of the numerator.
Then, we get $ = \dfrac{{a({{(\beta + \alpha )}^2} - 2\alpha \beta ) + b(\alpha + \beta )}}{{ab(\alpha + \beta ) + {a^2}\alpha \beta + {b^2}}}......(1)$
Now, we will put the values of $\alpha + \beta $and $\alpha \beta $ in equation $(1)$
After putting values we get: \[\dfrac{{a\left( {{{\left( {\dfrac{{ - b}}{a}} \right)}^2} - \dfrac{c}{a}} \right) + b\left( {\dfrac{{ - b}}{a}} \right)}}{{ab\left( {\dfrac{{ - b}}{a}} \right) + {a^2} \times \dfrac{c}{a} + {b^2}}}\]
We will simplify the above equation by expanding each term.
\[ = \dfrac{{{a}\left( {\dfrac{{{b^2} - 2ac}}{{{a} \times a}}} \right) + b\left( {\dfrac{{ - b}}{a}} \right)}}{{{a}b\left( {\dfrac{{ - b}}{{{a}}}} \right) + \dfrac{{{a} \times a \times c}}{{{a}}} + {b^2}}}\]
Here, we cancel out negative ${b^2}$ with positive ${b^2}$ and $a$ , $c$ present in numerator and denominator.
\[ = \dfrac{{\dfrac{{{{{b}}^2} - 2c{a} - {{{b}}^2}}}{{{a}}}}}{{ - {{{b}}^2} + ac + {{{b}}^2}}} = \dfrac{{ - 2{c}}}{{a{c}}}\]
$ \Rightarrow \dfrac{{ - 2}}{a}$. Finally, this is the result for above evaluation

Note: We should remember that we should simplify the equation and also put identities that are required to reach the point where we can put the values of sum and product of roots of the given equation.