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

\[ \frac{\beta}{a\alpha + b} + \frac{\alpha}{a\beta + b}. \]

Solution

Step 1: Write Relations Between Zeros and Coefficients

For the quadratic polynomial \( ax^2 + bx + c \),

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

Step 2: Combine the Fractions

\[ \frac{\beta}{a\alpha + b} + \frac{\alpha}{a\beta + b} = \frac{\beta(a\beta + b) + \alpha(a\alpha + b)} {(a\alpha + b)(a\beta + b)} \]

Step 3: Simplify the Numerator

\[ \beta(a\beta + b) + \alpha(a\alpha + b) = a(\alpha^2 + \beta^2) + b(\alpha + \beta) \]

Now,

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta \]

\[ = \left(-\frac{b}{a}\right)^2 – 2\left(\frac{c}{a}\right) = \frac{b^2}{a^2} – \frac{2c}{a} \]

So the numerator becomes:

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

\[ = \frac{b^2}{a} – 2c – \frac{b^2}{a} = -2c \]

Step 4: Simplify the Denominator

\[ (a\alpha + b)(a\beta + b) = a^2\alpha\beta + ab(\alpha + \beta) + b^2 \]

Substitute the values:

\[ = a^2\left(\frac{c}{a}\right) + ab\left(-\frac{b}{a}\right) + b^2 \]

\[ = ac – b^2 + b^2 = ac \]

Step 5: Find the Required Value

\[ \frac{\beta}{a\alpha + b} + \frac{\alpha}{a\beta + b} = \frac{-2c}{ac} \]

\[ = -\frac{2}{a} \]

Conclusion

The required value is:

\[ \boxed{-\frac{2}{a}} \]

\[ \therefore \quad \frac{\beta}{a\alpha + b} + \frac{\alpha}{a\beta + b} = -\frac{2}{a}. \]