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:

\[ \boxed{-\frac{25}{12}} \]

\[ \therefore \quad \frac{\alpha}{\beta} + \frac{\beta}{\alpha} = -\frac{25}{12}. \]

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