Quadratic Equation with Complex Roots: x²+4x+5=0
The problem: solve x² + 4x + 5 = 0.
Students find this problem unsettling because the familiar routine breaks halfway: the discriminant comes out negative, and suddenly the square root step asks for √(−4), which “does not exist” in the real numbers. Nothing is broken, though. A negative discriminant is the equation telling you something geometric: the parabola never touches the x-axis, so its roots cannot be real numbers. They are complex numbers instead, and the quadratic formula handles them with no extra machinery. Once you have seen one example worked all the way through, including the check, the fear disappears. This example shows every step.
Worked solution
We solve x² + 4x + 5 = 0, with a = 1, b = 4, c = 5.
Step 1. Compute the discriminant. D = b² − 4ac = 16 − 4(1)(5) = 16 − 20 = −4. Negative. That is the whole story in one number: two complex roots, no real roots. Do not stop here; the formula still works.
Step 2. Write the quadratic formula. x = (−b ± √(b² − 4ac)) / 2a, so:
x = (−4 ± √(−4)) / 2
Step 3. Simplify the square root of the negative number. √(−4) = √(4 · −1) = √4 · √(−1) = 2i, where i is the imaginary unit with i² = −1. So:
x = (−4 ± 2i) / 2
Step 4. Divide through. Split the fraction term by term:
x = −4/2 ± (2i)/2 = −2 ± i
The two roots are x = −2 + i and x = −2 − i, a conjugate pair. ✓
Step 5. Read the geometry. The parabola y = x² + 4x + 5 opens upward (a = 1 > 0) and its vertex sits at x = −b/2a = −2, where y = (−2)² + 4(−2) + 5 = 4 − 8 + 5 = 1. The lowest point of the parabola is ( −2, 1), one unit above the x-axis, so the curve floats entirely above the axis and never crosses it. That is exactly what “no real roots” means, drawn.
Why this method works
The quadratic formula is derived by completing the square, a purely algebraic process that never assumes the discriminant is positive. When b² − 4ac is negative, the square root step simply moves into the complex numbers, where every number has a square root. The ± then produces both roots, which are conjugates because the coefficients are real. The method does not change for complex roots; only the number system widens.
Common mistakes
- Declaring “no solution” at the negative discriminant. A negative discriminant means no real roots, not no roots at all. The equation still has two perfectly good complex solutions, −2 ± i.
- Writing √(−4) = −2. The square root of a negative number is imaginary, not negative-real. √(−4) = 2i (taking the principal branch); the ± in the formula, not the root, generates the pair.
- Forgetting to divide the imaginary part. Writing x = −4 ± 2i and stopping, skipping the /2. Every term in the numerator, real and imaginary alike, gets divided by 2a.
Check (never skip)
Plug x = −2 + i back into x² + 4x + 5. First, (−2 + i)² = 4 − 4i + i² = 4 − 4i − 1 = 3 − 4i. Next, 4(−2 + i) = −8 + 4i. Add everything: (3 − 4i) + (−8 + 4i) + 5 = (3 − 8 + 5) + (−4i + 4i) = 0 + 0i = 0. ✓ The root satisfies the equation exactly, imaginary parts canceling as they must. (The conjugate −2 − i checks the same way, with every i negated.)
Frequently Asked Questions
Q: What do complex roots mean geometrically?
A: The parabola never crosses the x-axis, so there is no real x with y = 0. The roots exist off the real line: at −2 + i and −2 − i in the complex plane. Algebraically they are full citizens; geometrically they are invisible on the ordinary xy-graph.
Q: Is √(−4) = 2i always correct?
A: With the standard principal branch of the square root, yes. The quadratic formula’s ± then produces both −2 + i and −2 − i automatically, which is why you never need a “±i” from the root itself.
Q: Do complex roots always come in pairs?
A: For any polynomial with real coefficients, yes: nonreal roots come in conjugate pairs a ± bi. If −2 + i is a root of x² + 4x + 5 = 0, then −2 − i must be one too, and the formula delivers both.
Try it with steps on our quadratic formula calculator.