Question:medium
Given that the equation $x^2 - (2a + b)x + \left(2a^2 + b^2 - b + \frac{1}{2}\right) = 0$ has two real roots. The value of $b$ is
Given that the equation $x^2 - (2a + b)x + \left(2a^2 + b^2 - b + \frac{1}{2}\right) = 0$ has two real roots. The value of $b$ is
Show Hint
When parameters are involved in discriminant problems, reduce everything to a perfect square to find maximum/minimum conditions easily.
Updated On: Apr 30, 2026
- $1$
- $2$
- $-1$
- $-2$
- $0$
Show Solution
The Correct Option is C
Solution and Explanation
Was this answer helpful?
0
Top Questions on Complex Numbers and Quadratic Equations
- If \( z = 3 + 4i \), then the modulus of \( z \) is:
- MHT CET - 2025
- Mathematics
- Complex Numbers and Quadratic Equations
- Find the roots of the quadratic equation \( 2x^2 - 4x - 6 = 0 \).
- MHT CET - 2025
- Mathematics
- Complex Numbers and Quadratic Equations
- The minimum value of \(| z+ \frac{3+4i}{2}|, |z| \leq 1\) is,
- JEE Main - 2024
- Mathematics
- Complex Numbers and Quadratic Equations
- \(If (\frac{(1+i}{1-i})^m=1, \text{then find the least positive integral value of m.}\)
- CBSE Class XI
- Mathematics
- Complex Numbers and Quadratic Equations
- Want to practice more? Try solving extra ecology questions todayView All Questions
