Value of an Expression Using Zeros of a Quadratic Polynomial

Video Explanation

Question

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

\[ p(s) = 3s^2 – 6s + 4, \]

find the value of

\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta. \]

Solution

Step 1: Write Relations Between Zeros and Coefficients

For a quadratic polynomial \( as^2 + bs + c \):

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

Here,

\[ a = 3,\quad b = -6,\quad c = 4 \]

\[ \alpha + \beta = 2, \quad \alpha\beta = \frac{4}{3} \]

Step 2: Find Each Part of the Expression

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

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = 4 – \frac{8}{3} = \frac{4}{3} \]

\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{4/3}{4/3} = 1 \]

Next,

\[ \frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha\beta} = \frac{2}{4/3} = \frac{3}{2} \]

Step 3: Substitute in the Given Expression

\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta \]

\[ = 1 + 2\left(\frac{3}{2}\right) + 3\left(\frac{4}{3}\right) \]

\[ = 1 + 3 + 4 \]

\[ = 8 \]

Conclusion

The required value is:

\[ \boxed{8} \]

\[ \therefore \quad \frac{\alpha}{\beta} + \frac{\beta}{\alpha} + 2\left(\frac{1}{\alpha} + \frac{1}{\beta}\right) + 3\alpha\beta = 8. \]