Question

The number of real values of \( m \) such that the equation \[ x^2 + (2m + 1)x + m = 0 \] has equal roots is:

• A. 1
• B. 0
• C. 2
• D. 3

Hint:

Always check the discriminant \( D = b^2 - 4ac \) when determining the nature of roots. If the result is negative, no real solutions exist.

Answer 1

The Correct Option is B

For a quadratic equation \( ax^2 + bx + c = 0 \) to have equal roots, the discriminant must be zero: \[ D = b^2 - 4ac = 0 \] Given: \[ \begin{align} a = 1, \quad b = 2m + 1, \quad c = m \Rightarrow D = (2m + 1)^2 - 4(1)(m) \Rightarrow D = 4m^2 + 4m + 1 - 4m = 4m^2 + 1 \] Set discriminant to zero: \[ 4m^2 + 1 = 0 \Rightarrow m^2 = -\frac{1}{4} \] This has no real solution since the square of a real number cannot be negative. Hence, there are no real values of \( m \) that make the roots equal.