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Quadratic Equations

A quadratic is any equation with an x² as its highest power. Learn every way to solve one — factoring, the square-root method, completing the square, and the quadratic formula — and how to read the parabola it graphs.

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Algebra Tutoring
The Main Ideas

Four Ways to Solve, One Shape to Know

A quadratic is any equation whose highest power is \(x^2\). Every one graphs as a parabola, and every one can be solved — factoring is fastest when it works, and the quadratic formula never fails.

Standard form

\[ ax^2 + bx + c = 0, \quad a \neq 0 \]

Get everything on one side equal to zero first — every method below assumes that.

Method 1: Factoring

\[ x^2 - 5x + 6 = 0 \Rightarrow (x-2)(x-3) = 0 \]

If it factors, set each factor to zero (zero-product property): \(x = 2\) or \(x = 3\). Fast, but not everything factors nicely.

Method 2: Square-root method

\[ x^2 = k \Rightarrow x = \pm\sqrt{k} \]

When there is no plain \(x\) term, isolate the square and take the root of both sides. Do not forget the \(\pm\).

Method 3: Completing the square

\[ x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2 \]

Rewrite as \((x + p)^2 = q\), then square-root. Always works, and it is where the quadratic formula comes from.

Method 4: The quadratic formula

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

Plug \(a\), \(b\), \(c\) straight in. This always works — reach for it whenever factoring is not obvious.

The discriminant tells you how many roots

\[ D = b^2 - 4ac \]
  • \(D > 0\) → two real solutions
  • \(D = 0\) → one (repeated) solution
  • \(D < 0\) → no real solutions

Reading the parabola

vertex

Opens up when \(a > 0\), down when \(a < 0\). The vertex sits on the axis of symmetry at \(x = -\tfrac{b}{2a}\); the real solutions are where the parabola crosses the x-axis.

See It Worked Out

Worked Examples

Example 1 · Factoring

Solve: \(x^2 - 5x + 6 = 0\)

  1. Two numbers that multiply to \(+6\) and add to \(-5\): \(-2\) and \(-3\).

    \[ (x - 2)(x - 3) = 0 \]

  2. Set each factor to zero.

    \[ x - 2 = 0 \ \text{or}\ x - 3 = 0 \]
\( x = 2 \ \text{or}\ x = 3 \)

Example 2 · Square-root method

Solve: \(2x^2 - 50 = 0\)

  1. Isolate the square term.

    \[ 2x^2 = 50 \Rightarrow x^2 = 25 \]

  2. Take the square root of both sides — keep the \(\pm\).

    \[ x = \pm\sqrt{25} \]
\( x = 5 \ \text{or}\ x = -5 \)

Example 3 · Quadratic formula

Solve: \(x^2 + 4x + 1 = 0\)

  1. Identify \(a = 1\), \(b = 4\), \(c = 1\) and plug into the formula.

    \[ x = \frac{-4 \pm \sqrt{4^2 - 4(1)(1)}}{2(1)} \]

  2. Simplify under the root: \(16 - 4 = 12\), and \(\sqrt{12} = 2\sqrt{3}\).

    \[ x = \frac{-4 \pm 2\sqrt{3}}{2} \]

  3. Divide every term by 2.

    \[ x = -2 \pm \sqrt{3} \]
\( x = -2 + \sqrt{3} \ \text{or}\ x = -2 - \sqrt{3} \)
Your Turn

Practice Problems

Solve each equation, then reveal the solution to check yourself.

\(x^2 - 9 = 0\)

Show solution

Square-root method: \(x^2 = 9\).

\[ x = \pm 3 \]

\(x^2 + 7x + 10 = 0\)

Show solution

Factors: \(+2\) and \(+5\) multiply to 10 and add to 7.

\[ (x + 2)(x + 5) = 0 \Rightarrow x = -2 \ \text{or}\ x = -5 \]

\(x^2 - 6x + 9 = 0\)

Show solution

Perfect-square trinomial; the discriminant is 0, so one repeated root.

\[ (x - 3)^2 = 0 \Rightarrow x = 3 \]

\(3x^2 - 12 = 0\)

Show solution

Isolate: \(3x^2 = 12 \Rightarrow x^2 = 4\).

\[ x = \pm 2 \]

\(x^2 + 2x - 4 = 0\)

Show solution

Does not factor nicely — use the formula with \(a=1, b=2, c=-4\). Discriminant \(= 4 + 16 = 20\), \(\sqrt{20} = 2\sqrt{5}\).

\[ x = \frac{-2 \pm 2\sqrt{5}}{2} = -1 \pm \sqrt{5} \]
Keep Going

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