Roots of Quadratic Equation — Types and Formulas

Roots of Quadratic Equation — Types and Formulas

TL;DR

The roots of a quadratic equation ax² + bx + c = 0 are the values of x that satisfy it. The discriminant D = b² − 4ac classifies them into three types — real and distinct, real and equal, or complex. This article covers the quadratic formula, Vieta's sum-and-product relations, three worked examples, and a 1,400-year history from Brahmagupta to Cardano.

A Curve Crosses an Axis — and That's a Root

The graph of y = ax² + bx + c is a parabola. The roots of the quadratic equation ax² + bx + c = 0 are the x-values where this parabola crosses the x-axis. A parabola can cross the x-axis twice, touch it once, or miss it entirely — those three cases correspond to the three types of roots.

The roots of a quadratic equation are also called its zeros or its solutions — three names for the same idea. A quadratic equation has at most two roots, because its degree is 2.

The Quadratic Formula

For the standard-form quadratic ax² + bx + c = 0 with a ≠ 0, the two roots are:

x = \frac{-b \pm \sqrt{b² - 4ac}}{2a}.

The expression under the square root — b² − 4ac — is the discriminant, written D or Δ. The discriminant alone tells you what kind of roots the equation has, without computing them.

Discriminant value Nature of roots Geometric meaning
D > 0 Two real and distinct roots Parabola crosses x-axis at two points
D = 0 One real root (repeated, "double root") Parabola touches x-axis at exactly one point (the vertex)
D < 0 Two complex conjugate roots, no real roots Parabola never crosses or touches the x-axis

The discriminant is also what you check first in any quadratic-equation problem — if the question only asks for the nature of the roots, you can answer it without computing the roots themselves.

Vieta's Sum and Product of Roots

For ax² + bx + c = 0 with roots r₁ and r₂:

r₁ + r₂ = −b/a, r₁ ⋅ r₂ = c/a.

These are Vieta's formulas. They are derived by expanding a(x − r₁)(x − r₂) = ax² + bx + c and matching coefficients. The practical use: if a problem gives you the sum or product of roots and asks you to construct a quadratic, Vieta's formulas hand you the coefficients in one step.

A quadratic with roots 3 and -2, for instance, has sum 1 and product -6. Taking a = 1, the quadratic is x² − 1 ⋅ x + (-6) = x² − x − 6 = 0.

Methods of Finding Roots — Side by Side

Method When it works best When it fails
Factoring a, b, c are small integers; roots are rational Roots irrational or complex — no clean factorisation
Quadratic formula Always works Slower for clean factorable cases
Completing the square When you need vertex form along the way Slower than the formula for pure root-finding
Graphing Visual approximation; finding integer roots quickly Exact irrational or complex roots
Sum/product (Vieta's) Constructing a quadratic from given roots Not for finding unknown roots

The default first pass is factoring; if the factors aren't immediate within ~30 seconds, switch to the quadratic formula. Completing the square stays in the curriculum because it derives the quadratic formula and connects to vertex form.

Three Worked Examples — Quick, Standard, Stretch

Quick. Find the roots of x² − 7x + 12 = 0.

Try factoring first. Two numbers that multiply to 12 and add to -7: that's -3 and -4.

So:

x² − 7x + 12 = (x - 3)(x - 4) = 0.

Apply the zero product property: x = 3 or x = 4.

Final answer: Roots are x = 3 and x = 4. Both real and distinct.

Quick discriminant check: D = (-7)² − 4(1)(12) = 49 − 48 = 1 > 0. Two real distinct roots — matches.

Standard (Wrong Path First — The Detour Students Take). Find the roots of 2x² + 3x − 5 = 0.

The wrong path. A student tries to factor: "two numbers that multiply to 2⋅(-5) = -10 and add to 3..." After three tries the student lands on 5 and -2 — which sum to 3 and multiply to -10. The student writes "(x + 5)(x - 2) = 0" and tries to verify...

The student's instinct was right — but it has an extra rewriting step the student skipped.

The clean rescue. Skip the factoring tangle. Apply the quadratic formula with a = 2, b = 3, c = −5:

x = \frac{-3 \pm \sqrt{3² - 4(2)(-5)}}{2(2)}.

So x = 1 or x = -\frac{5}{2}.

Final answer: Roots are x = 1 and x = -\frac{5}{2}.

Stretch. Find the roots of x² − 2x + 5 = 0.

Discriminant first: D = (-2)² − 4(1)(5) = 4 - 20 = -16. Negative discriminant means no real roots; the roots are complex conjugates.

Apply the formula:

x = 1 ± 2i.

Final answer: Roots are x = 1 + 2i and x = 1 - 2i.

Why Roots of Quadratics Matter

Quadratic roots are the most-used solution set in early algebra. They surface everywhere a quantity grows quadratically or where a balance of two opposing effects produces a turning point.

The Mathematicians Who Shaped the Quadratic

The quadratic formula has a 1,400-year history of incremental refinements.

Common Errors When Working With Roots of Quadratic Equations

Mistake 1: Misreading the sign of b.

Correct way: b is the signed coefficient.

Mistake 2: Forgetting the ± in the formula.

Correct way: A quadratic with D > 0 has two real roots.

Mistake 3: Confusing "no real roots" with "no roots at all."

Correct way: D < 0 means there are no real roots, but still two complex roots.

Conclusion

Sharpen Your Roots — Three Practice Problems

  1. Find the roots of x² − 9x + 20 = 0 by factoring.
  2. Find the roots of 2x² + 4x + 5 = 0. (What type of roots?)
  3. Construct a quadratic equation whose roots are −3 and 7, using Vieta's formulas.