Value of an Expression Using Zeros of a Quadratic Polynomial
Video Explanation
Question
If \( \alpha \) and \( \beta \) are the zeros of the quadratic polynomial
\[ f(x) = 6x^2 + x – 2, \]
find the value of
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha}. \]
Solution
Step 1: Use Relations Between Zeros and Coefficients
For a quadratic polynomial \( ax^2 + bx + c \):
\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]
Here,
\[ a = 6,\quad b = 1,\quad c = -2 \]
\[ \alpha + \beta = -\frac{1}{6}, \quad \alpha\beta = -\frac{1}{3} \]
Step 2: Simplify the Required Expression
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^2 + \beta^2}{\alpha\beta} \]
Now,
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta \]
Step 3: Substitute the Values
\[ \alpha^2 + \beta^2 = \left(-\frac{1}{6}\right)^2 – 2\left(-\frac{1}{3}\right) \]
\[ = \frac{1}{36} + \frac{2}{3} = \frac{25}{36} \]
\[ \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{25/36}{-1/3} \]
\[ = -\frac{25}{12} \]
Conclusion
The required value is:
\[ \boxed{-\frac{25}{12}} \]
\[ \therefore \quad \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{25}{12}. \]