Quadratic Polynomial from Given Zeros

Video Explanation

Question

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

\[ f(x) = x^2 – 1, \]

find the quadratic polynomial whose zeros are

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

Solution

Step 1: Find Sum and Product of \( \alpha \) and \( \beta \)

Given polynomial:

\[ x^2 – 1 = 0 \]

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

\[ a = 1,\quad b = 0,\quad c = -1 \]

\[ \alpha + \beta = -\frac{b}{a} = 0 \]

\[ \alpha\beta = \frac{c}{a} = -1 \]

Step 2: Find the Sum of the New Zeros

Let the new zeros be \[ \frac{2\alpha}{\beta} \text{ and } \frac{2\beta}{\alpha}. \]

Sum of new zeros:

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

Now,

\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta = 0 – 2(-1) = 2 \]

So,

\[ \frac{2\alpha}{\beta} + \frac{2\beta}{\alpha} = 2\left(\frac{2}{-1}\right) = -4 \]

Step 3: Find the Product of the New Zeros

\[ \left(\frac{2\alpha}{\beta}\right)\left(\frac{2\beta}{\alpha}\right) = 4 \]

Step 4: Form the Required Quadratic Polynomial

If the sum of zeros is \(S\) and the product is \(P\), the quadratic polynomial is:

\[ x^2 – Sx + P \]

Here,

\[ S = -4,\quad P = 4 \]

So the required quadratic polynomial is:

\[ x^2 + 4x + 4 \]

Conclusion

The required quadratic polynomial is:

\[ \boxed{x^2 + 4x + 4} \]

\[ \therefore \quad x^2 + 4x + 4 \text{ is the required polynomial.} \]