Question

If \(\alpha,\beta\) are the zeroes of the quadratic polynomial \(ax^2 + bx+c\), \(a ≠0\) then \(\alpha^2+\beta^2=\)

• A. \(\frac{1}{a^2}(b^2+2ac)\)
• B. \(\frac{1}{a^2}(c^2+2ab)\)
• C. \(\frac{1}{a^2}(b^2-2ac)\)
• D. \(\frac{1}{a^2}(c^2-2ab)\)

Answer 1

The Correct Option is C

Step 1: Use the identity: 
\( \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha\beta \)

Step 2: From the quadratic equation \( ax^2 + bx + c = 0 \), we know:
- Sum of the roots \( \alpha + \beta = -\dfrac{b}{a} \)
- Product of the roots \( \alpha \beta = \dfrac{c}{a} \)

Step 3: Substitute in the identity:
\[ \alpha^2 + \beta^2 = \left( \dfrac{-b}{a} \right)^2 - 2 \cdot \dfrac{c}{a} = \dfrac{b^2}{a^2} - \dfrac{2c}{a} \]

Step 4: Take LCM:
\[ \alpha^2 + \beta^2 = \dfrac{b^2 - 2ac}{a^2} \]

The correct option is (C): \(\frac{1}{a^2}(b^2-2ac)\)