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Completing the Square

Completing the Square – definition and example, Algebra 1 math glossary – Mathovore

Key takeaway: Completing the square rewrites a quadratic so that one side is a perfect square, by adding (b/2)² to both sides, which lets you solve with a square root.

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

Definition of Completing the Square

Definition

Completing the square is a method for solving quadratic equations that do not factor neatly. You take \(x^{2} + bx\) and add exactly the number that turns it into a perfect square trinomial, which can then be written as \((x + \frac{b}{2})^{2}\).

An area picture explains the name. The expression \(x^{2} + 6x\) is a square of side \(x\) plus two rectangles of size \(3 \times x\). The corner needed to complete a bigger square is a \(3 \times 3\) square of area 9. After adding that corner to both sides, take the square root of each side, remembering both the positive and negative roots. The same steps convert standard form to vertex form.

Formula and notation

The key step is adding the square of half the middle coefficient:

\[x^{2} + bx + (\frac{b}{2})^{2} = (x + \frac{b}{2})^{2}\]

This only works directly when the leading coefficient is 1; otherwise divide the equation by \(a\) first.

x²3x3x9x3x² + 6x + 9 = (x + 3)²the missing corner is 9(6 ÷ 2)² = 9
The dashed corner of area 9 completes the square of side x + 3.

Key facts

  • Move the constant to the other side before you complete the square.
  • Whatever number you add to one side must also be added to the other side.
  • Remember the plus-or-minus sign when taking the square root.
  • The method always works, and it proves the quadratic formula.

Completing the Square example

Problem

Use completing the square to solve \(x^{2} + 6x – 11 = 0\), giving exact and rounded answers.

  1. Move the constant across: \(x^{2} + 6x = 11\).

  2. Half of 6 is 3, and \(3^{2} = 9\). Add 9 to both sides: \(x^{2} + 6x + 9 = 20\), so \((x + 3)^{2} = 20\).

  3. Take roots: \(x + 3 = \pm\sqrt{20} = \pm 2\sqrt{5}\), so \(x = -3 \pm 2\sqrt{5}\).

The solutions are \(x = -3 + 2\sqrt{5} \approx 1.47\) and \(x = -3 – 2\sqrt{5} \approx -7.47\).

Common mistake

Watch out

Dina writes \(x^{2} + 6x + 9 = 11\), adding 9 to the left side only, and then gets \((x + 3)^{2} = 11\).

Why it is wrong

An equation is a balance. Adding 9 on the left means 9 must be added on the right as well: \(11 + 9 = 20\). Her version solves a different equation. With the correct right side, \((x + 3)^{2} = 20\).

Related terms and lessons

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

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