Proof Using Zeros of a Quadratic Polynomial

Video Explanation

Question

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

\[ f(x) = x^2 – px + q, \]

prove that

\[ \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{p^4}{q^2} – \frac{4p^2}{q} + 2. \]

Solution

Step 1: Write 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,

\[ a = 1,\quad b = -p,\quad c = q \]

\[ \alpha + \beta = p, \quad \alpha\beta = q \]

Step 2: Start with the LHS

\[ \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{\alpha^4 + \beta^4}{\alpha^2\beta^2} \]

Step 3: Simplify the Numerator

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

Also,

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

\[ = p^2 – 2q \]

So,

\[ \alpha^4 + \beta^4 = (p^2 – 2q)^2 – 2q^2 \]

\[ = p^4 – 4p^2q + 4q^2 – 2q^2 \]

\[ = p^4 – 4p^2q + 2q^2 \]

Step 4: Substitute in the Expression

\[ \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{p^4 – 4p^2q + 2q^2}{q^2} \]

\[ = \frac{p^4}{q^2} – \frac{4p^2}{q} + 2 \]

Conclusion

Hence,

\[ \frac{\alpha^2}{\beta^2} + \frac{\beta^2}{\alpha^2} = \frac{p^4}{q^2} – \frac{4p^2}{q} + 2 \]

\[ \therefore \quad \text{The given identity is proved.} \]