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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-
a(α2β+β2α)+b(αβ+βα)\mathrm a\left(\dfrac{\mathrm\alpha^2}{\mathrm\beta}+\dfrac{\mathrm\beta^2}{\mathrm\alpha}\right)+\mathrm b\left(\dfrac{\mathrm\alpha}{\mathrm\beta}+\dfrac{\mathrm\beta}{\mathrm\alpha}\right)

Explanation

Solution

This is a question of quadratic equations. First we will take the LCMs of the two terms involving a and b. We will use the relationship that the sum of roots of quadratic equations, α+β\alpha + \beta , can be written as −ba - \dfrac{b}{a}. Also, the product of the roots αβ\alpha \beta can be written as ca\dfrac{c}{a}. We will then open the brackets and write the roots in terms of their sum and product. We can simplify them further using the formulas.

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-
a(α2β+β2α)+b(αβ+βα)aˉ(α3+β3βα)+b(α2+β2βα)aˉ((α+β)(α2+β2−αβ)αβ)+b((α+β)2−2αβαβ)aˉ((α+β)((α+β)2−3αβ)αβ)+b((α+β)2−2αβαβ)\mathrm a\left(\dfrac{\mathrm\alpha^2}{\mathrm\beta}+\dfrac{\mathrm\beta^2}{\mathrm\alpha}\right)+\mathrm b\left(\dfrac{\mathrm\alpha}{\mathrm\beta}+\dfrac{\mathrm\beta}{\mathrm\alpha}\right)\\\=\mathrm a\left(\dfrac{\mathrm\alpha^3+\mathrm\beta^3}{\mathrm{\mathrm\beta\mathrm\alpha}}\right)+\mathrm b\left(\dfrac{\mathrm\alpha^2+\mathrm\beta^2}{\mathrm{\mathrm\beta\mathrm\alpha}}\right)\\\=\mathrm a\left(\dfrac{\left(\mathrm\alpha+\mathrm\beta\right)\left(\mathrm\alpha^2+\mathrm\beta^2-\mathrm{\mathrm\alpha\mathrm\beta}\right)}{\mathrm{\mathrm\alpha\mathrm\beta}}\right)+\mathrm b\left(\dfrac{\left(\mathrm\alpha+\mathrm\beta\right)^2-2\mathrm{\mathrm\alpha\mathrm\beta}}{\mathrm{\mathrm\alpha\mathrm\beta}}\right)\\\=\mathrm a\left(\dfrac{\left(\mathrm\alpha+\mathrm\beta\right)\left(\left(\mathrm\alpha+\mathrm\beta\right)^2-3\mathrm{\mathrm\alpha\mathrm\beta}\right)}{\mathrm{\mathrm\alpha\mathrm\beta}}\right)+\mathrm b\left(\dfrac{\left(\mathrm\alpha+\mathrm\beta\right)^2-2\mathrm{\mathrm\alpha\mathrm\beta}}{\mathrm{\mathrm\alpha\mathrm\beta}}\right)
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)((−ba)2−3ca)ca)+b((−ba)2−2caca)aˉ((−b)(b2−3ac)a2c)+b(b2−2acac)−b3+3abcacˉ+b3−2abcacabcacˉ=b=\mathrm a\left(\dfrac{\left({\displaystyle\dfrac{-\mathrm b}{\mathrm a}}\right)\left(\left({\displaystyle\dfrac{-\mathrm b}{\mathrm a}}\right)^2-{\displaystyle\dfrac{3\mathrm c}{\mathrm a}}\right)}{\displaystyle\dfrac{\mathrm c}{\mathrm a}}\right)+\mathrm b\left(\dfrac{\left({\displaystyle\dfrac{-\mathrm b}{\mathrm a}}\right)^2-{\displaystyle\dfrac{2\mathrm c}{\mathrm a}}}{\displaystyle\dfrac{\mathrm c}{\mathrm a}}\right)\\\=\mathrm a\left(\dfrac{\left(-\mathrm b\right)\left(\mathrm b^2-3\mathrm{ac}\right)}{\mathrm a^2\mathrm c}\right)+\mathrm b\left(\dfrac{\mathrm b^2-2\mathrm{ac}}{\mathrm{ac}}\right)\\\=\dfrac{-\mathrm b^3+3\mathrm{abc}}{\mathrm{ac}}+\dfrac{\mathrm b^3-2\mathrm{abc}}{\mathrm{ac}}\\\=\dfrac{\mathrm{abc}}{\mathrm{ac}}=\mathrm b
This is the required answer.

Note:
The above given formula signifies that the sum of roots of a quadratic equation is the negative of the ratio of coefficient of xx and x2x^2. Also, the product of roots is the ratio of constant term and coefficient of x2x^2. The most common mistake is that the students often forget the negative sign in the formula for the sum of roots, which often leads to the wrong answer.