Question

If \( \alpha, \beta \) are the zeroes of the quadratic polynomial \( px^2 + qx + r \), then find the value of \( \alpha^3\beta + \beta^3\alpha \).

Hint:

Express any symmetric function of roots in terms of $(\alpha + \beta)$ and $(\alpha\beta)$ to solve polynomial relation problems easily.

Answer 1

Step 1: Understanding the Concept:
For a quadratic polynomial \( ax^2 + bx + c \), the sum of zeroes is \( \alpha + \beta = -b/a \) and the product of zeroes is \( \alpha\beta = c/a \).
Step 2: Key Formula or Approach:
From the given polynomial \( px^2 + qx + r \):
Sum of zeroes \( \alpha + \beta = -\frac{q}{p} \)
Product of zeroes \( \alpha\beta = \frac{r}{p} \)
Step 3: Detailed Explanation:
We need to evaluate the expression:
\[ E = \alpha^3\beta + \beta^3\alpha \]
Taking common factor \( \alpha\beta \):
\[ E = \alpha\beta(\alpha^2 + \beta^2) \]
Using the identity \( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \):
\[ E = \alpha\beta [(\alpha + \beta)^2 - 2\alpha\beta] \]
Substituting the values of sum and product:
\[ E = \left(\frac{r}{p}\right) \left[ \left(-\frac{q}{p}\right)^2 - 2\left(\frac{r}{p}\right) \right] \]
\[ E = \left(\frac{r}{p}\right) \left[ \frac{q^2}{p^2} - \frac{2r}{p} \right] \]
Taking LCM inside the bracket:
\[ E = \left(\frac{r}{p}\right) \left[ \frac{q^2 - 2pr}{p^2} \right] \]
\[ E = \frac{r(q^2 - 2pr)}{p^3} \]
Step 4: Final Answer:
The value is \( \frac{r(q^2 - 2pr)}{p^3} \).