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: Write the equation in standard form.
The given equation is \(x^2 + 9 = 0\), which is already in the standard form \(ax^2 + bx + c = 0\) with \(a = 1\), \(b = 0\), \(c = 9\).
Step 2: Calculate the discriminant.
The discriminant is \(D = b^2 - 4ac = 0^2 - 4(1)(9) = 0 - 36 = -36\).
Step 3: Interpret the discriminant.
Since \(D = -36 < 0\), the discriminant is negative, which means the roots are not real. They are complex (imaginary) numbers.
Step 4: Confirm by solving directly.
From \(x^2 = -9\), we get \(x = \pm\sqrt{-9} = \pm 3i\), which are indeed imaginary.
Step 5: Eliminate other options.
Real and equal roots require \(D = 0\); real and opposite roots would require \(D > 0\) and \(b = 0\); rational roots require \(D\) to be a perfect square and non-negative. None of these apply here.
Step 6: Select the correct answer.
The roots are not real, corresponding to option 2.
\[ \boxed{\text{not real}} \]