Quadratic Polynomial from Given Zeros

Video Explanation

Question

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

\[ f(x) = x^2 – 3x – 2, \]

find a quadratic polynomial whose zeros are

\[ \frac{1}{2\alpha + \beta} \quad \text{and} \quad \frac{1}{2\beta + \alpha}. \]

Solution

Step 1: Find \( \alpha + \beta \) and \( \alpha\beta \)

Comparing \(x^2 – 3x – 2\) with \(ax^2 + bx + c\),

\[ a = 1,\quad b = -3,\quad c = -2 \]

\[ \alpha + \beta = -\frac{b}{a} = 3 \]

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

Step 2: Find the Sum of the New Zeros

\[ \frac{1}{2\alpha + \beta} + \frac{1}{2\beta + \alpha} = \frac{(2\beta + \alpha) + (2\alpha + \beta)} {(2\alpha + \beta)(2\beta + \alpha)} \]

\[ = \frac{3(\alpha + \beta)} {2(\alpha^2 + \beta^2) + 5\alpha\beta} \]

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = 9 + 4 = 13 \]

Denominator:

\[ 2(13) + 5(-2) = 26 – 10 = 16 \]

\[ \text{Sum of new zeros} = \frac{3 \times 3}{16} = \frac{9}{16} \]

Step 3: Find the Product of the New Zeros

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

\[ = \frac{1}{16} \]

Step 4: Form the Required Quadratic Polynomial

Required polynomial:

\[ x^2 – (\text{sum})x + (\text{product}) \]

\[ = x^2 – \frac{9}{16}x + \frac{1}{16} \]

Multiplying throughout by 16:

\[ 16x^2 – 9x + 1 \]

Conclusion

The required quadratic polynomial is:

\[ \boxed{16x^2 – 9x + 1} \]

\[ \therefore \quad 16x^2 – 9x + 1 \text{ is the required polynomial.} \]