1
Negative discriminant means no REAL solution
It does not mean no solution. x² + 4 = 0 has no real solution and two complex ones, 2i and -2i. Dropping the word "real" is the error the whole lesson is aimed at.
Lesson · Quadratics and Complex Numbers · Complex numbers
Complex Roots of Quadratics
When b² - 4ac is negative the quadratic formula still works: the root of the negative discriminant becomes a multiple of i, and the two solutions are conjugates of one another. Every quadratic with real coefficients has exactly two solutions once complex numbers are allowed.
b² - 4ac < 0 → two solutions, neither of them real
x = -b/(2a) ± ( √(4ac - b²) / (2a) ) i
the pair is p + qi and p - qi: conjugates, since only ± differs
sum = -b/a product = c/a both still real
What a negative discriminant produces
Core ideas
What this lesson establishes
“The Discriminant” counted the real solutions. This lesson finds the ones it counted as absent.
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The two roots are conjugates
The formula produces -b/(2a) ± (something)i, and the ± is the only difference between the two answers. So one is p + qi and the other is p - qi, provided a, b, c are real.
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A conjugate pair is a parabola that misses the axis
x-intercepts are real solutions. If there are none of those, the curve never reaches the x-axis — the two solutions are still there, but not at any point of the picture.
The material
Solving with a negative discriminant
The formula does not change. What changes is what is under the root.
“The Discriminant” read the sign of D = b² - 4ac and reported two, one or no real solutions. The last case was left open on purpose: √D with D < 0 had no meaning yet. “Complex Numbers” gave it one, and nothing else about the method needs adjusting.
Theorem · Solutions when the discriminant is negative
Let a, b, c be real with a ≠ 0, and let D = b² - 4ac < 0. Then ax² + bx + c = 0 has exactly two solutions, x = -b/(2a) ± (√(-D)/(2a)) i, and each is the conjugate of the other.
D < 0, so -D > 0 and √(-D) is an ordinary positive real number. Since (i√(-D))² = i² · (-D) = -(-D) = D, the quantity i√(-D) is a square root of D.
Putting that into the formula gives x = (-b ± i√(-D)) / (2a), and dividing both parts of the numerator by 2a gives x = -b/(2a) ± (√(-D)/(2a)) i.
a, b and D are real and a ≠ 0, so -b/(2a) and √(-D)/(2a) are both real numbers. The two solutions therefore have the same real part and opposite imaginary parts, which is what it means to be conjugates.
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Both parts of the numerator are divided by 2a. Writing x = (-6 ± 4i)/2 as -3 ± 4i divides the real part and forgets the other; the correct reduction is -3 ± 2i.
Example · x² - 6x + 13 = 0
D = 36 - 52 = -16, so √(-D) = √16 = 4 and x = (6 ± 4i)/2 = 3 ± 2i. The two solutions are 3 + 2i and 3 - 2i.
The same pair by completing the square
x² - 6x + 13 = 0
(x - 3)² - 9 + 13 = 0 half of -6 is -3, squared is 9
(x - 3)² = -4
x - 3 = ±2i since (2i)² = -4
x = 3 ± 2i
Two methods, computed independently, agreeing. That is the pattern the whole course is built on, and it is worth noticing that completing the square never had to mention the discriminant: the negative appears on its own, as the number a square is being asked to equal.
Theorem · Conjugate pairs need real coefficients
The pairing is a consequence of a, b and c being real numbers. Without that hypothesis it fails: x² - ix = 0 factors as x(x - i) = 0, so its solutions are 0 and i, which are not conjugates of one another.
With the pair in hand the quadratic factors, just not over the reals: x² - 6x + 13 = (x - 3 - 2i)(x - 3 + 2i). Expanding confirms it, since (x - 3)² - (2i)² = x² - 6x + 9 + 4. Polynomials and Factoring called such a quadratic irreducible; that word always means irreducible over some system, and over the complex numbers no quadratic is.
The sum and product survive as a check. For 3 ± 2i the sum is 6 = -b/a and the product is (3 + 2i)(3 - 2i) = 9 + 4 = 13 = c/a. Both come out real, as they must: the imaginary parts cancel in the sum and the conjugate product is real by “Operations with Complex Numbers”.
Example · 2x² + 3x + 5 = 0
D = 9 - 40 = -31, which is negative and not a perfect square. x = (-3 ± i√31)/4 = -3/4 ± (√31/4) i. The root √31 stays exactly as it is; there is nothing to simplify and a decimal would only lose information.
On the graph nothing dramatic happens. The parabola y = x² - 6x + 13 sits entirely above the x-axis — its lowest point is (3, 4) — so it has no x-intercepts, which is the picture of "no real solutions". The conjugate pair is not hiding anywhere in that picture, and “Graphs of Quadratic Functions” explains why the picture only ever shows real numbers.
Interactive
The roots a negative discriminant hides
The conjugate pair, multiplied back into the quadratic it came from
Method
Solving when the discriminant is negative
The discriminant is computed first, so that nothing later is a surprise.
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Compute D = b² - 4ac before anything else
If it is negative you already know the shape of the answer: two conjugate solutions, and no x-intercept on the graph.
2
Write the root of the negative as a multiple of i
√-36 = 6i, because (6i)² = 36i² = -36. Take the positive multiple here; the ± already in the formula supplies the other sign, and writing -6i as well would count each solution twice.
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Divide BOTH parts of the numerator by 2a
(-2 ± 6i)/4 is -1/2 ± (3/2)i. This is the step that is quietly skipped more often than any other in the lesson.
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Check one solution by substituting it
The other needs no separate check. Conjugating an equation with real coefficients conjugates both sides, and 0 is its own conjugate, so if one of the pair works the other does too.
Negative discriminants
Score 0 / 0Choose the answer you can justify.
2x² + 2x + 5 = 0
a = 2, b = 2, c = 5. Nothing here factors and there is no missing linear term, so the formula is the method.
D = b² - 4ac = 2² - 4(2)(5) = 4 - 40 = -36 negative
x = ( -2 ± √-36 ) / (2·2)
√-36 = 6i since (6i)² = 36i² = -36
x = ( -2 ± 6i ) / 4
2 6 1 3
x = - -- ± -- i = - -- ± -- i
4 4 2 2
check x = -1/2 + (3/2)i :
x² = 1/4 - (3/2)i + (9/4)i² = -2 - (3/2)i
2x² = -4 - 3i
2x = -1 + 3i
2x² + 2x + 5 = (-4 - 3i) + (-1 + 3i) + 5 = 0 ✓
Both parts of the numerator were divided by 4. Stopping at (-2 ± 6i)/4, or dividing only the -2, are the two ways this last step fails, and the substitution above catches either of them immediately.
The other solution is -1/2 - (3/2)i. Its check is the conjugate of the one written out, line for line, so doing it again would confirm nothing new.
For a faded complex-root solve, use 3x² - 6x + 10 = 0. The supplied discriminant is D = -84. Convert its root to 2i√21, reduce the whole formula and use the sum and product as independent checks. You should reach x = 1 ± (√21/3)i, whose sum is 2 = -b/a and whose product is 10/3 = c/a. A remaining denominator of 6 means the common factor 2 was not removed from both numerator terms.
Common mistakes
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Answering "no solution" with no qualifierAfter this lesson the honest answers are "no real solution" or "two complex solutions". x² + 4 = 0 has both descriptions and they do not contradict each other.
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Dividing only part of the numerator by 2a(-2 ± 6i)/4 is -1/2 ± (3/2)i, not -1/2 ± 6i. The fraction bar runs under the whole numerator, exactly as it does when the discriminant is positive.
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Writing √-36 as -6(-6)² = 36, not -36, so -6 is not a square root of -36. The minus does not come out of a square root as a factor of -1; it comes out as a factor of i, and i is not a real number that can be absorbed into the sign.
Check your understanding
✓
Finish when a negative discriminant makes you reach for i rather than stop.
Take any quadratic with D < 0, produce the conjugate pair in the form p ± qi with p and q exact, and verify one of them by substitution. If the check is a nuisance, the arithmetic of “Operations with Complex Numbers” needs another pass before this one does.
Note: Every quadratic with real coefficients now has exactly two solutions — two distinct real ones, one repeated real one, or a conjugate pair. That completeness is what the complex numbers were introduced for. The remaining lessons return to real coefficients and real graphs, where the discriminant decides how the parabola meets the axis.