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Question: If \(\mathrm\alpha\;\mathrm{and}\;\mathrm\beta\) are the zeroes of the quadratic polynomial \(f(x) =...

If α  and  β\mathrm\alpha\;\mathrm{and}\;\mathrm\beta are the zeroes of the quadratic polynomial f(x)=ax2+bx+cf(x) = ax^2+ bx + c, then evaluate the following-
βaα+b+ααβ+b\dfrac{\mathrm\beta}{\mathrm{a\alpha}+\mathrm b}+\dfrac{\mathrm\alpha}{\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm b}

Explanation

Solution

This is a question of quadratic equations. We will compute the LCM of the two denominators and then take the common for the terms in a and b, which will enable us to alter the expression in terms of the sum or roots and product of roots. Also, in these formulas, we must always ensure that the coefficient of x2x^2 is 1. Then we will use the formulas-
α+β=−baαβ=ca\mathrm\alpha+\mathrm\beta=-\dfrac{\mathrm b}{\mathrm a}\\\\\mathrm{\mathrm\alpha\mathrm\beta}=\dfrac{\mathrm c}{\mathrm a}

Complete step-by-step answer :
We have to convert the given expression such that it can be expressed in the form of α+β  and  αβ\mathrm\alpha+\mathrm\beta\;\mathrm{and}\;\mathrm{\mathrm\alpha\mathrm\beta} only, so that we can apply the given formulas, substitute the values and find the result.
So we will convert the equation as follows-
βaα+b+ααβ+bβ(αβ+b)+α(aα+b)(aα+b)(αβ+b)ˉαβ2+bβ+aα2+bαa2αβ+ab(α+β)+b2ˉa(α2+β2)+b(α+β)a2αβ+ab(α+β)+b2ˉa((α+β)2−2αβ)+b(α+β)a2αβ+ab(α+β)+b2ˉ\dfrac{\mathrm\beta}{\mathrm{a\alpha}+\mathrm b}+\dfrac{\mathrm\alpha}{\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm b}\\\=\dfrac{\mathrm\beta\left(\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm b\right)+\mathrm\alpha\left(\mathrm{a\alpha}+\mathrm b\right)}{\left(\mathrm{a\alpha}+\mathrm b\right)\left(\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm b\right)}\\\=\dfrac{\mathrm{\mathrm\alpha\mathrm\beta}^2+\mathrm{b\beta}+\mathrm{a\alpha}^2+\mathrm{b\alpha}}{\mathrm a^2\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{ab}\left(\mathrm\alpha+\mathrm\beta\right)+\mathrm b^2}\\\=\dfrac{\mathrm a\left(\mathrm\alpha^2+\mathrm\beta^2\right)+\mathrm b\left(\mathrm\alpha+\mathrm\beta\right)}{\mathrm a^2\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{ab}\left(\mathrm\alpha+\mathrm\beta\right)+\mathrm b^2}\\\=\dfrac{\mathrm a\left(\left(\mathrm\alpha+\mathrm\beta\right)^2-2\mathrm{\mathrm\alpha\mathrm\beta}\right)+\mathrm b\left(\mathrm\alpha+\mathrm\beta\right)}{\mathrm a^2\mathrm{\mathrm\alpha\mathrm\beta}+\mathrm{ab}\left(\mathrm\alpha+\mathrm\beta\right)+\mathrm b^2}
Now we will substitute the given formula in this expression to find its value using-
α+β=−baαβ=ca\mathrm\alpha+\mathrm\beta=-\dfrac{\mathrm b}{\mathrm a}\\\\\mathrm{\mathrm\alpha\mathrm\beta}=\dfrac{\mathrm c}{\mathrm a}
=a((−ba)2−2ca)+b(−ba)a2(ca)+ab(−ba)+b2b2a−2c−b2aac−b2+b2ˉ−2cacˉ=  −2a=\dfrac{\mathrm a\left(\left({\displaystyle\dfrac{-\mathrm b}{\mathrm a}}\right)^2-{\displaystyle\dfrac{2\mathrm c}{\mathrm a}}\right)+\mathrm b\left({\displaystyle\dfrac{-\mathrm b}{\mathrm a}}\right)}{\mathrm a^2\left({\displaystyle\dfrac{\mathrm c}{\mathrm a}}\right)+\mathrm{ab}\left({\displaystyle\dfrac{-\mathrm b}{\mathrm a}}\right)+\mathrm b^2}\\\=\dfrac{{\displaystyle\dfrac{\mathrm b^2}{\mathrm a}}-2\mathrm c-{\displaystyle\dfrac{\mathrm b^2}{\mathrm a}}}{\mathrm{ac}-\mathrm b^2+\mathrm b^2}\\\=\dfrac{-2\mathrm c}{\mathrm{ac}}=\;\dfrac{-2}{\mathrm a}
This is the required answer.

Note :The most common mistake here is that the students take the LCM incorrectly, and do not multiply the terms accordingly to the numerator. Another common mistake is that students get confused between the terms a, b and α,β\alpha, \beta and often exchange them. We should always apply the formula after completely simplifying the initial equation.