Question

The roots of the quadratic equation \(x^2 + 9 = 0\) are

• A. real and equal
• B. not real
• C. real and negative of each other
• D. rational numbers

Hint:

For any equation of the form \(x^2 + k = 0\) where \(k \gt 0\), the roots are always imaginary (not real) because \(x^2 = -k\) requires taking the square root of a negative number.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks about the nature of the roots of the quadratic equation \(x^2 + 9 = 0\).
We need to determine whether the roots are real and equal, not real, real and opposite in sign, or rational numbers.

Step 2: Key Formula or Approach:
For a standard quadratic equation \(ax^2 + bx + c = 0\), the nature of the roots is determined by the discriminant (\(D\)): \[ D = b^2 - 4ac \] - If \(D \gt 0\), the roots are real and distinct.
- If \(D = 0\), the roots are real and equal.
- If \(D \lt 0\), the roots are not real (imaginary).

Step 3: Detailed Explanation:
1. Identify the coefficients of the given quadratic equation \(x^2 + 9 = 0\):
Here, we have: \[ a = 1 \] \[ b = 0 \] \[ c = 9 \] 2. Calculate the value of the discriminant \(D\): \[ D = b^2 - 4ac \] \[ D = (0)^2 - 4(1)(9) \] \[ D = 0 - 36 \] \[ D = -36 \] 3. Since the discriminant \(D = -36\), which is less than 0 (\(D \lt 0\)), the quadratic equation has no real roots.
Alternatively, solving directly for \(x\): \[ x^2 = -9 \] \[ x = \pm \sqrt{-9} = \pm 3i \] Since the square root of a negative number is not a real number, the roots are complex/imaginary.

Step 4: Final Answer:
The discriminant of the equation is negative, which means the roots of the given quadratic equation are not real.
Hence, the correct option is (B).