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Discriminant

Discriminant – definition and example, Algebra 1 math glossary – Mathovore

Key takeaway: The discriminant is the expression b² − 4ac from a quadratic equation, and its sign tells whether the equation has two real solutions, one, or none.

Grade 9 – Algebra 1A-REI.B.4bExpressions and Equations

Definition of Discriminant

Definition

The discriminant is the part of the quadratic formula that sits under the square root. Calculating it before solving tells you what kind of answers to expect, which saves work and catches mistakes.

If the value is positive, the square root is a real number and the plus and minus signs give two different solutions, so the parabola crosses the x-axis twice. If it equals zero, the root is zero, there is one repeated solution, and the parabola just touches the axis at its vertex. If it is negative, no real number is a square root of it, so there are no real solutions and the curve stays entirely on one side of the axis. A perfect-square positive value also signals that the quadratic factors with whole numbers.

Formula and notation

For \(ax^{2} + bx + c = 0\) the discriminant is

\[D = b^{2} – 4ac\]

The outcome: \(D > 0\) gives two real solutions, \(D = 0\) gives one, and \(D < 0\) gives none. Use parentheses when \(b\) or \(c\) is negative.

D > 0two solutionsD = 0one solutionD < 0no real solutions
Positive: two crossings. Zero: one touch. Negative: no contact with the x-axis.

Key facts

  • One number built from a, b and c summarizes the whole solution count.
  • A zero discriminant means the quadratic is a perfect square trinomial.
  • A negative discriminant means the graph never meets the x-axis.
  • It can be checked in seconds, before attempting any other solving method.

Discriminant example

Problem

Compute the discriminant of each equation and tell how many real solutions it has: (a) \(3x^{2} + 4x – 4 = 0\), (b) \(4x^{2} – 20x + 25 = 0\), (c) \(x^{2} + 2x + 7 = 0\).

  1. (a) \(D = 4^{2} – 4(3)(-4) = 16 + 48 = 64 > 0\): two solutions.

  2. (b) \(D = (-20)^{2} – 4(4)(25) = 400 – 400 = 0\): one repeated solution, \(x = \frac{5}{2}\).

  3. (c) \(D = 2^{2} – 4(1)(7) = 4 – 28 = -24 < 0\): no real solutions.

The counts are 2, 1 and 0 real solutions, respectively.

Common mistake

Watch out

For \(3x^{2} + 4x – 4 = 0\), Ana computes \(16 – 4(3)(4) = -32\) and declares that the equation has no real solutions.

Why it is wrong

She dropped the negative sign of \(c = -4\). The product \(4ac = 4(3)(-4) = -48\), and subtracting a negative number adds: \(16 + 48 = 64\). A positive value means two solutions, and since 64 is a perfect square the equation factors as \((3x – 2)(x + 2) = 0\).

Related terms and lessons

Browse the glossary: A–Z list · Algebra 1 vocabulary · Expressions and Equations

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