Lesson · Quadratics and Complex Numbers · Solving quadratics

The Discriminant

b² - 4ac is the quantity under the radical in the quadratic formula, and its sign decides how many real solutions there are: two, one or none. Computing it first costs one line and tells you what kind of answer to expect, or that there is nothing to look for.

Open the interactive lesson Practice

D = b² - 4ac D > 0 two distinct real solutions D = 0 exactly one real solution, x = -b/(2a) D < 0 no real solution

One number, three cases

Core ideas

What this lesson establishes

Nothing new is computed here. The discriminant is a piece of the formula, given a name and read on its own.

The sign is the whole message

D sits under a square root, and a square root of a negative number is not real. That is the entire reason there are three cases, and it is why the size of D matters far less than its sign.

D = 0 means one solution, not none

The formula becomes x = (-b ± 0)/(2a), and both branches give -b/(2a). So the equation has exactly one solution and the quadratic is a constant times a perfect square. "No solutions" is the D < 0 case, and the two are not neighbours.

For integer coefficients, a perfect square means it factors

If a, b, c are integers and D is the square of an integer, then √D is an integer and both solutions are rational, so the quadratic factors over the rationals. That turns "should I try factoring?" into a calculation instead of a guess.

The material

Reading the answer before finding it

The definition, the three cases with their proofs, and the extra thing a perfect square tells you.

Definition · Discriminant

For ax² + bx + c = 0 with a ≠ 0, the discriminant is D = b² - 4ac.

It is the quantity under the radical in the quadratic formula, and it is computed from the coefficients of the standard form, signs included.

Theorem · What the sign of D decides

Let a, b, c be real with a ≠ 0, and let D = b² - 4ac. If D > 0 then ax² + bx + c = 0 has two distinct real solutions. If D = 0 it has exactly one. If D < 0 it has none.

“The Quadratic Formula” showed the equation is equivalent to (x + b/(2a))² = D/(4a²), and 4a² is positive, so the right-hand side has the same sign as D.

If D > 0 the square root property gives two solutions, x = -b/(2a) ± √D/(2a). They are distinct because √D is not 0 and 2a is not 0.

If D = 0 the equation reads (x + b/(2a))² = 0, whose only solution is x = -b/(2a).

If D < 0 the right-hand side is negative, and no real number has a negative square. There is no real solution.

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The three cases are exhaustive and they do not overlap, so the discriminant always answers the question, and never answers it twice.

What D = 0 looks like

4x² - 12x + 9 = 0 has D = 144 - 144 = 0, and its factorisation is (2x - 3)². The two facts travel together: D = 0 exactly when the quadratic is a constant times a perfect square, and then the single solution is the double root of “Solving by Factoring”.

Perfect squares and rational solutions

Theorem · When a quadratic with integer coefficients factors

Let a, b and c be integers with a ≠ 0. The equation ax² + bx + c = 0 has rational solutions if and only if D = b² - 4ac is the square of an integer.

Suppose D = m² for some integer m ≥ 0. Then √D = m, and the solutions (-b ± m)/(2a) are quotients of integers with a non-zero denominator, so both are rational.

Conversely, suppose some solution x is rational. Multiplying (x + b/(2a))² = D/(4a²) through by 4a² gives (2ax + b)² = D, so D is the square of the rational number 2ax + b. A rational number whose square is an integer is itself an integer — the same fact Polynomials and Factoring's rational root theorem rests on — so 2ax + b is an integer and D is the square of one.

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So the search of “Factoring Simple Trinomials” in Polynomials and Factoring can be called off before it starts. x² - 4x + 1 = 0 has D = 16 - 4 = 12, which is not a perfect square, so no rational factorisation exists and no amount of looking will produce one. x² - 5x + 6 = 0 has D = 1, so one does, and it is (x - 2)(x - 3).

What the discriminant does not tell you

It gives the number of real solutions, and for integer coefficients whether they are rational. It does not give the solutions, it does not say whether they are positive or negative, and its size carries no message about difficulty: D = 1 and D = 10000 both mean "two real solutions".

One derived fact is worth keeping. The two solutions are -b/(2a) ± √D/(2a), so they differ by √D/|a|. The gap depends on a as well: with a held fixed, a larger D puts the two answers further apart, which is a statement about the graph in “Graphs of Quadratic Functions” rather than about the solving.

Example · Four discriminants

x² - 5x + 6: D = 25 - 24 = 1, positive and a perfect square, so two rational solutions, 2 and 3. 4x² - 12x + 9: D = 0, so one solution, 3/2. x² - 4x + 1: D = 12, positive and not a perfect square, so two irrational solutions, 2 ± √3. x² + x + 1: D = -3, so no real solution.

Each of the four took one line to classify, and three of them took longer than that to solve.

The habit worth building is to compute D on every quadratic before choosing a method. A perfect square sends you to factoring, a D of 0 tells you to expect a single answer, and a negative D ends the work at this point in the course.

Interactive

The discriminant

Counting the roots before finding them

Method

Using the discriminant first

One line of arithmetic, then a decision about which method to use.

Reach standard form and read a, b, c with their signs

The same first step as “The Quadratic Formula”, for the same reason. A D computed from the wrong coefficients answers a question about a different equation.

Compute D with brackets

For 2x² - 4x - 3 = 0, D = (-4)² - 4(2)(-3) = 16 + 24 = 40, not 16 - 24. With a positive, a negative c makes -4ac positive, and this is where the sign goes missing.

Read the case off the sign

Positive gives two real solutions, zero gives one, negative gives none — and in the last case the answer is already complete.

If D is positive and the coefficients are integers, test for a perfect square

If it is one, factoring will work and is probably quicker. If it is not, go straight to the formula and expect an irrational answer.

Counting before solving

Score 0 / 0

Choose the answer you can justify.

Four quadratics, classified then solved

D is computed on the first line each time, and the method is chosen from it.

A. x² - 5x + 6 = 0 a = 1, b = -5, c = 6 D = 25 - 24 = 1 positive, and 1 = 1² → two rational roots factor: (x - 2)(x - 3) = 0 x = 2 or x = 3 B. 4x² - 12x + 9 = 0 a = 4, b = -12, c = 9 D = 144 - 144 = 0 → exactly one solution x = -b/(2a) = 12/8 = 3/2 and indeed 4x² - 12x + 9 = (2x - 3)² C. x² - 4x + 1 = 0 a = 1, b = -4, c = 1 D = 16 - 4 = 12 positive, not a perfect square → two irrational roots, no factorisation x = (4 ± √12)/2 = (4 ± 2√3)/2 = 2 ± √3 D. x² + x + 1 = 0 a = 1, b = 1, c = 1 D = 1 - 4 = -3 negative → no real solution

C is the one that pays. Polynomials and Factoring's product-and-sum search over the divisors of 1 would have been tried and would have failed, and D = 12 says in advance that it must: 12 is not a perfect square, so no rational factorisation exists. One line replaced a search.

D was settled by a single subtraction. “Complex Numbers”, “Operations with Complex Numbers” and “Complex Roots of Quadratics” supply the numbers and method that continue beyond the reals. Nothing in this lesson changes when those roots arrive — D < 0 will still mean "no real solution", which is why the word real is worth writing now.

For a faded method-choice pass, classify these before doing any full solution: x² - 11x + 24 = 0, 5x² - 45 = 0, x² + 8x = 3, and 2x² + x + 4 = 0. The supplied cues are, in order, an integer factor pair, no linear term, an expression one constant away from a square, and a negative discriminant. Choose factoring, the square root property, completing the square, and stopping at no real solution respectively; then solve the first three and use substitution to check the roots.

Common mistakes

Dropping the sign in -4acWhen a and c have opposite signs, -4ac is positive and D grows. For 2x² - 4x - 3 = 0, D = 16 + 24 = 40, not 16 - 24 = -8 — the difference between two real solutions and none at all.

Reading D = 0 as "no solutions"D = 0 gives exactly one solution, -b/(2a). It is the boundary between two and none, and it belongs to neither of them.

Saying "no solutions" when "no real solutions" is meantBefore “Complex Numbers” the two phrases look interchangeable, and they are not. x² + 1 = 0 has no real solution and two complex ones. The sentence you write today should still be true three lessons from now.

Check your understanding

✓

Finish when D is the first thing you compute on any quadratic.

Given the coefficients you should produce D, name the case, say whether the answers will be rational, and only then pick a method. Getting the sign of -4ac right when c is negative is the single skill that makes the whole habit reliable.

Note: The discriminant is the last thing this course does with real numbers alone. D < 0 is not the end of the story: “Complex Numbers” defines a number whose square is -1, “Complex Roots of Quadratics” returns to every equation this lesson dismissed, and “Graphs of Quadratic Functions” shows what a negative discriminant looks like — a parabola that misses the x-axis entirely.