Algebra

Quadratic equations

Factoring, completing the square and the quadratic formula — and what the discriminant tells you.

About 11 minutes

A quadratic equation has the form ax^2 + bx + c = 0 with a \ne 0. Its graph is a parabola, and its solutions are the points where the parabola crosses the x-axis — which can happen twice, once, or not at all.

Solving by factoring

If the quadratic factors as (x - p)(x - q) = 0, then one of the factors must be zero, so x = p or x = q. For x^2 + bx + c, look for two numbers with product $c$ and sum $b$ .

x^{2} - (p + q)x + pq = (x - p)(x - q)

Completing the square

Any quadratic can be rewritten as a perfect square plus a constant. This always works, and it is where the quadratic formula comes from. It also reveals the vertex of the parabola.

ax^{2} + bx + c = a\left(x + \frac{b}{2a}\right)^{2} + c - \frac{b^{2}}{4a}

The quadratic formula

Completing the square on the general equation gives a formula that solves every quadratic.

The quantity under the root, \Delta = b^2 - 4ac, is the discriminant. If \Delta > 0 there are two real solutions; if \Delta = 0 there is one repeated solution; if \Delta < 0 there are no real solutions, only a pair of complex conjugates.

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

Worked examples

Solve x^2 - 5x + 6 = 0.

Two numbers with product 6 and sum $-5$ : $-2$ and $-3$ .
Factor: (x - 2)(x - 3) = 0.
So x = 2 or x = 3.

Answer: x = 2 or x = 3

Solve 2x^2 + 3x - 2 = 0.

a = 2, b = 3, c = -2.
\Delta = 3^2 - 4(2)(-2) = 25.
x = \frac{-3 \pm 5}{4}.
So x = \frac12 or x = -2.

Answer: x = \frac{1}{2} or x = -2

Common mistakes

Dividing both sides by $x$ and losing the solution x = 0.
Applying the formula before moving every term to one side.
Sign slips in $-b$ and in -4ac when $b$ or $c$ is negative.
Dividing only the root by $2a$ instead of the whole numerator.

Practice problems

1.Solve x^2 - 9 = 0
2.Solve x^2 + 4x + 4 = 0
3.Solve x^2 - x - 1 = 0
4.How many real solutions does x^2 + 2x + 5 = 0 have?

Practise this topic with checked answers

Tools for this lesson

Quadratic Equation SolverSolve quadratic equations with the discriminant and quadratic formula shown step by step, including exact radical and complex roots.Completing the SquareRewrite a quadratic in vertex form a(x − h)² + k with every step, and read off the vertex, axis of symmetry and extreme value.Expression FactorizerFactor polynomials and algebraic expressions over the rationals, with the result checked against the original expression.Graphing CalculatorPlot and compare functions, implicit equations and inequalities with zoom, pan, coordinate tracking and PNG export.

Next lesson: The Pythagorean theorem

Solving by factoring

If the quadratic factors as

(x - p)(x - q) = 0

(x - p)(x - q) = 0, then one of the factors must be zero, so

x = p

x = p or

x = q

x = q. For

x^2 + bx + c

x^2 + bx + c, look for two numbers with product

c

and sum

b

x^{2} - (p + q)x + pq = (x - p)(x - q)

x^{2} - (p + q)x + pq = (x - p)(x - q)

Completing the square

Any quadratic can be rewritten as a perfect square plus a constant. This always works, and it is where the quadratic formula comes from. It also reveals the vertex of the parabola.

ax^{2} + bx + c = a\left(x + \frac{b}{2a}\right)^{2} + c - \frac{b^{2}}{4a}

ax^{2} + bx + c = a\left(x + \frac{b}{2a}\right)^{2} + c - \frac{b^{2}}{4a}

The quadratic formula

Completing the square on the general equation gives a formula that solves every quadratic.

The quantity under the root,

\Delta = b^2 - 4ac

\Delta = b^2 - 4ac, is the discriminant. If

\Delta > 0

\Delta > 0 there are two real solutions; if

\Delta = 0

\Delta = 0 there is one repeated solution; if

\Delta < 0

\Delta < 0 there are no real solutions, only a pair of complex conjugates.

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

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

Worked examples

Solve $x^2 - 5x + 6 = 0$ x^2 - 5x + 6 = 0.

Two numbers with product 6 and sum $-5$ : $-2$ and $-3$ .
Factor: $(x - 2)(x - 3) = 0$ (x - 2)(x - 3) = 0.
So $x = 2$ x = 2 or $x = 3$ x = 3.

Answer: $x = 2$ x = 2 or $x = 3$ x = 3

Solve $2x^2 + 3x - 2 = 0$ 2x^2 + 3x - 2 = 0.

$a = 2$ a = 2, $b = 3$ b = 3, $c = -2$ c = -2.
$\Delta = 3^2 - 4(2)(-2) = 25$ \Delta = 3^2 - 4(2)(-2) = 25.
$x = \frac{-3 \pm 5}{4}$ x = \frac{-3 \pm 5}{4}.
So $x = \frac12$ x = \frac12 or $x = -2$ x = -2.

Answer: $x = \frac{1}{2}$ x = \frac{1}{2} or $x = -2$ x = -2

Practice problems

1.Solve

x^2 - 9 = 0

x^2 - 9 = 0

2.Solve

x^2 + 4x + 4 = 0

x^2 + 4x + 4 = 0

3.Solve

x^2 - x - 1 = 0

x^2 - x - 1 = 0

4.How many real solutions does

x^2 + 2x + 5 = 0

x^2 + 2x + 5 = 0 have?

Practise this topic with checked answers