Evaluation of \( \alpha^4 + \beta^4 \)
Video Explanation
Question
If \( \alpha \) and \( \beta \) are the zeroes of the quadratic polynomial
\[ f(x) = ax^2 + bx + c, \]
evaluate
\[ \alpha^4 + \beta^4. \]
Solution
Step 1: Write Relations Between Zeros and Coefficients
For the quadratic polynomial \( ax^2 + bx + c \),
\[ \alpha + \beta = -\frac{b}{a}, \quad \alpha\beta = \frac{c}{a} \]
Step 2: Express \( \alpha^4 + \beta^4 \) in Terms of Lower Powers
\[ \alpha^4 + \beta^4 = (\alpha^2 + \beta^2)^2 – 2\alpha^2\beta^2 \]
Step 3: Find \( \alpha^2 + \beta^2 \)
\[ \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha\beta \]
\[ = \left(-\frac{b}{a}\right)^2 – 2\left(\frac{c}{a}\right) = \frac{b^2 – 2ac}{a^2} \]
Step 4: Substitute in the Expression
\[ \alpha^4 + \beta^4 = \left(\frac{b^2 – 2ac}{a^2}\right)^2 – 2\left(\frac{c}{a}\right)^2 \]
\[ = \frac{(b^2 – 2ac)^2 – 2a^2c^2}{a^4} \]
Conclusion
The required value is:
\[ \boxed{ \alpha^4 + \beta^4 = \frac{(b^2 – 2ac)^2 – 2a^2c^2}{a^4} } \]
\[ \therefore \quad \alpha^4 + \beta^4 = \dfrac{(b^2 – 2ac)^2 – 2a^2c^2}{a^4}. \]