Step-by-step guide with 6 worked examples and 4 practice problems
Completing the square is an algebraic technique that rewrites a quadratic expression \( ax^2 + bx + c \) into vertex form \( a(x - h)^2 + k \). It is used to find the vertex of a parabola, solve quadratic equations (it is the method behind the quadratic formula), and prepare certain expressions for integration in calculus.
The simplest case uses the identity obtained by halving the middle coefficient, squaring it, and adding/subtracting it:
\[ x^2 + bx = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 \]For the general quadratic \( ax^2 + bx + c \) with \( a \ne 0 \), factor out \( a \) from the \(x\)-terms, then complete the square inside the parentheses:
\[ ax^2 + bx + c = a\left(x + \frac{b}{2a}\right)^2 + \left(c - \frac{b^2}{4a}\right) \]| Case | Formula |
|---|---|
| Simple: \(x^2 + bx\) | \( x^2 + bx = \left(x + \tfrac{b}{2}\right)^2 - \tfrac{b^2}{4} \) |
| General: \(ax^2 + bx + c\) | \( ax^2 + bx + c = a\left(x + \tfrac{b}{2a}\right)^2 + \left(c - \tfrac{b^2}{4a}\right) \) |
Vertex form. The general form above is exactly the vertex form \( a(x - h)^2 + k \) with \( h = -b/(2a) \) and \( k = c - b^2/(4a) \). The vertex of the parabola \( y = ax^2 + bx + c \) is at \( (h, k) = \left(-b/(2a),\, c - b^2/(4a)\right) \).
Step 1. The coefficient of \(x\) is \(b = 4\). Half of \(b\) is \(2\), and \(2^2 = 4\).
Step 2. Add and subtract \(4\):
\[ \begin{aligned} x^2 + 4x &= \left(x + \frac{4}{2}\right)^2 - \left(\frac{4}{2}\right)^2 \\ &= (x + 2)^2 - 4 \end{aligned} \]Check: \((x + 2)^2 - 4 = x^2 + 4x + 4 - 4 = x^2 + 4x\) ✓
Step 1. Focus on \(x^2 + 2x\). Here \(b = 2\), half is \(1\), and \(1^2 = 1\).
Step 2. Complete the square on \(x^2 + 2x\), keep the \(+5\) outside:
\[ \begin{aligned} x^2 + 2x + 5 &= \left[\left(x + \frac{2}{2}\right)^2 - \left(\frac{2}{2}\right)^2\right] + 5 \\ &= (x + 1)^2 - 1 + 5 \\ &= (x + 1)^2 + 4 \end{aligned} \]Vertex: \((-1,\, 4)\).
Step 1. Factor \(2\) out of the \(x\)-terms:
\[ 2x^2 - 12x = 2(x^2 - 6x) \]Step 2. Complete the square inside the parentheses. Here \(b = -6\), half is \(-3\), and \((-3)^2 = 9\):
\[ \begin{aligned} 2x^2 - 12x &= 2\left[\left(x - 3\right)^2 - 9\right] \\ &= 2(x - 3)^2 - 18 \end{aligned} \]Vertex: \((3,\, -18)\).
Step 1. Factor \(-1\) out of the \(x\)-terms:
\[ -x^2 - 10x = -(x^2 + 10x) \]Step 2. Complete the square inside. Here \(b = 10\), half is \(5\), and \(5^2 = 25\):
\[ \begin{aligned} -x^2 - 10x &= -\left[(x + 5)^2 - 25\right] \\ &= -(x + 5)^2 + 25 \end{aligned} \]Vertex: \((-5,\, 25)\).
Step 1. Factor \(-2\) out of the \(x\)-terms:
\[ -2x^2 - 3x = -2\left(x^2 + \tfrac{3}{2}x\right) \]Step 2. Complete the square inside. Here \(b = 3/2\), half is \(3/4\), and \((3/4)^2 = 9/16\):
\[ \begin{aligned} -2x^2 - 3x &= -2\left[\left(x + \tfrac{3}{4}\right)^2 - \tfrac{9}{16}\right] \\ &= -2\left(x + \tfrac{3}{4}\right)^2 + 2 \cdot \tfrac{9}{16} \\ &= -2\left(x + \tfrac{3}{4}\right)^2 + \tfrac{9}{8} \end{aligned} \]Vertex: \(\left(-\tfrac{3}{4},\, \tfrac{9}{8}\right)\).
Step 1. Factor \(-3\) out of the \(x\)-terms, keep \(+2\) outside:
\[ -3x^2 + 2x + 2 = -3\left(x^2 - \tfrac{2}{3}x\right) + 2 \]Step 2. Complete the square inside. Here \(b = -2/3\), half is \(-1/3\), and \((-1/3)^2 = 1/9\):
\[ \begin{aligned} -3x^2 + 2x + 2 &= -3\left[\left(x - \tfrac{1}{3}\right)^2 - \tfrac{1}{9}\right] + 2 \\ &= -3\left(x - \tfrac{1}{3}\right)^2 + 3 \cdot \tfrac{1}{9} + 2 \\ &= -3\left(x - \tfrac{1}{3}\right)^2 + \tfrac{1}{3} + 2 \\ &= -3\left(x - \tfrac{1}{3}\right)^2 + \tfrac{7}{3} \end{aligned} \]Vertex: \(\left(\tfrac{1}{3},\, \tfrac{7}{3}\right)\).
Try completing the square on each expression, then check your work.
Here \(b = -6\), half is \(-3\), square is \(9\).
\[ \begin{aligned} x^2 - 6x + 2 &= (x - 3)^2 - 9 + 2 \\ &= (x - 3)^2 - 7 \end{aligned} \]Here \(b = 6\), half is \(3\), square is \(9\).
\[ \begin{aligned} x^2 + 6x - 3 &= (x + 3)^2 - 9 - 3 \\ &= (x + 3)^2 - 12 \end{aligned} \]Completing the square on the general equation \(ax^2 + bx + c = 0\) is exactly how the quadratic formula is derived:
An algebraic technique that rewrites a quadratic \(ax^2 + bx + c\) into vertex form \(a(x - h)^2 + k\) by adding and subtracting the square of half the coefficient of \(x\).
Simple case: \(x^2 + bx = (x + b/2)^2 - (b/2)^2\). General case: \(ax^2 + bx + c = a(x + b/(2a))^2 + (c - b^2/(4a))\).
First factor \(a\) out of the \(x^2\) and \(x\) terms: \(ax^2 + bx + c = a(x^2 + (b/a)x) + c\). Then complete the square inside the parentheses using \(b/a\) as the new middle coefficient, and distribute the outer factor \(a\) back through the perfect square and the subtracted constant.
Starting from \(ax^2 + bx + c = 0\) and dividing by \(a\), completing the square on the left gives \((x + b/(2a))^2 = (b^2 - 4ac)/(4a^2)\). Taking square roots and isolating \(x\) yields the quadratic formula.
Last updated: October 20, 2026