Evaluation of an Expression Using Zeros of a Quadratic Polynomial

Video Explanation

Question

If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial

\[ f(x) = ax^2 + bx + c, \]

evaluate

\[ a\!\left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right) + b\!\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right). \]

Solution

Step 1: Relations Between Zeros and Coefficients

For \( ax^2 + bx + c \),

\[ \alpha + \beta = -\frac{b}{a}, \qquad \alpha\beta = \frac{c}{a} \]

Step 2: Rewrite Each Term

\[ \frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha} = \frac{\alpha^3+\beta^3}{\alpha\beta}, \qquad \frac{\alpha}{\beta}+\frac{\beta}{\alpha} = \frac{\alpha^2+\beta^2}{\alpha\beta} \]

Use identities:

\[ \alpha^3+\beta^3=(\alpha+\beta)^3-3\alpha\beta(\alpha+\beta), \quad \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta \]

Step 3: Substitute the Values

Let \( S=\alpha+\beta=-\dfrac{b}{a} \) and \( P=\alpha\beta=\dfrac{c}{a} \).

\[ a\left(\frac{S^3-3PS}{P}\right) + b\left(\frac{S^2-2P}{P}\right) \]

Substituting \(S\) and \(P\):

\[ a\left(\frac{-\dfrac{b^3}{a^3}+\dfrac{3bc}{a^2}}{\dfrac{c}{a}}\right) + b\left(\frac{\dfrac{b^2}{a^2}-\dfrac{2c}{a}}{\dfrac{c}{a}}\right) \]

Simplifying:

\[ \left(-\frac{b^3}{ac}+3b\right) +\left(\frac{b^3}{ac}-2b\right) \]

\[ = b \]

Conclusion

The required value is:

\[ \boxed{b} \]

\[ \therefore \quad a\!\left(\frac{\alpha^2}{\beta}+\frac{\beta^2}{\alpha}\right) + b\!\left(\frac{\alpha}{\beta}+\frac{\beta}{\alpha}\right) = b. \]