How to Solve Quadratic Equations by Completing the Square

Completing the square rewrites a quadratic such as x² + 6x + 8 into the form (x + p)² + q, which lets you solve equations exactly and read a parabola’s vertex straight off the equation. This Edexcel IGCSE Maths guide works through the method step by step, for a = 1 and for a larger coefficient, then uses it to solve and to graph, with worked examples and free auto-marked practice questions below.

Prior Knowledge This page requires confidence with expanding double brackets and solving linear equations.
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Core Ideas of Completing the Square

Start with the Quadratic
Begin with the standard form \( ax^2 + bx + c \).
Halve it
Take half of the x-coefficient \( \tfrac{b}{2} \).
Square it
Square that half to get \( \bigl(\tfrac{b}{2}\bigr)^2 \).
Take it away
Add and subtract the same square to form a perfect square bracket.

What does "Completing the Square" produce?

We aim to write a quadratic as \( a(x+p)^2 + q \). For solving, arrange to \( (x+p)^2 = q \). When \( a\ne1 \), factor \( a \) from the \( x^2 \) and \( x \) terms only and complete the square inside the bracket.

Completing the Square: Step by Step

Each step is shown applied to \(x^2 + 6x + 8\).

  1. Halve the x-coefficient. Divide the coefficient of \(x\) by 2.
    \(6 \div 2 = 3\)

    This gives the number that goes inside the bracket with \(x\).

  2. Write the squared bracket. Use the halved value inside \((x + \square)^2\). Expanding it shows the hidden constant.
    \((x + 3)^2 = x^2 + 6x + 9\)

    The bracket reproduces the \(x^2\) and \(6x\) terms, but it also creates a \(+9\).

  3. Subtract the square you added. \((x+3)^2\) contains a \(+9\) that wasn't in the original; subtract it straight away.
    \((x + 3)^2 - 9\)

    This keeps the expression equivalent to the original \(x^2 + 6x\).

  4. Bring in the original constant. Add the constant term from the original quadratic to finish.
    \((x + 3)^2 - 9 + 8 = (x + 3)^2 - 1\)

    Combine \(-9 + 8 = -1\) to get the final completed-square form.

General result (\(a = 1\)) \[x^2 + bx + c = \left(x + \tfrac{b}{2}\right)^2 - \left(\tfrac{b}{2}\right)^2 + c\]
When \(a \ne 1\): add one step before starting. Factor \(a\) out of the \(x^2\) and \(x\) terms only: \(ax^2 + bx + c = a\!\left(x^2 + \tfrac{b}{a}x\right) + c\). Then apply the four steps above to the expression inside the bracket, and finally multiply \(a\) back through the constants.

Worked Examples

💡 Example 1: Solve by completing the square

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

\[ \begin{aligned} x^2+6x+5&=0\\[4pt] x^2+6x&=-5\\[4pt] x^2+6x+9&=-5+9\\[4pt] (x+3)^2&=4\\[4pt] x+3&=\pm2\\[4pt] x&=-3\pm2\\[4pt] x&=-1,\,-5 \end{aligned} \]
What's happening?

Move the constant to the right-hand side.

Halve the x-coefficient: \(6 \div 2 = 3\). Square it: \(3^2 = 9\). Add 9 to both sides.

The left side is now a perfect square.

Square-root both sides (remember \(\pm\)).

Subtract 3 from each value to find the two solutions.

💡 Example 2: \( a\ne1 \), factor first

Solve \( 2x^2 + 8x + 6 = 0 \).

\[ \begin{aligned} 2x^2+8x+6&=0\\[4pt] 2(x^2+4x)+6&=0\\[4pt] 2\!\left((x+2)^2-4\right)+6&=0\\[4pt] 2(x+2)^2-8+6&=0\\[4pt] 2(x+2)^2-2&=0\\[4pt] (x+2)^2&=1\\[4pt] x+2&=\pm1\\[4pt] x&=-1,\,-3 \end{aligned} \]
What's happening?

Factor \(2\) from the \(x^2\) and \(x\) terms only; the \(+6\) stays outside.

Inside the bracket: halve \(4 \to 2\), square \(2 \to 4\). Write \((x+2)^2 - 4\).

Expand the \(2\) across the constants: \(2 \times (-4) = -8\).

Combine: \(-8 + 6 = -2\). Then isolate \((x+2)^2\).

Square-root both sides and solve.

Using Completing the Square for Graphing

Write \( ax^2+bx+c \) as \( a(x+p)^2+q \). The vertex is \((-p,\,q)\); axis of symmetry \( x=-p \). If \( a>0 \) the graph opens up; if \( a<0 \) it opens down.

x y -2 -1 1 2 3 4 5 5 4 3 2 1 -1 -2 -3 -4 0 x = 2 (2, -3) (0, 1) y = (x - 2)² - 3
The completed-square form y = (x - 2)² - 3 reveals the vertex at (2, -3) and the axis of symmetry x = 2 directly from the equation.

💡 Example 3: Rewrite for graphing

\[ \begin{aligned} y&=x^2-4x+1\\ &=(x^2-4x+4)-4+1\\ &=(x-2)^2-3 \end{aligned} \]

Vertex \((2,-3)\); axis \(x=2\); opens up since \(a=1>0\).

💡 Example 4: \( a\ne1 \)

\[ \begin{aligned} y&=-3x^2+12x+5\\ &=-3(x^2-4x)+5\\ &=-3\!\left((x-2)^2-4\right)+5\\ &=-3(x-2)^2+17 \end{aligned} \]

Vertex \((2,17)\); axis \(x=2\); opens down since \(a=-3<0\).

🔑 Key Points

  • Mantra: halve it, square it, take it away.
  • When \( a\ne1 \), factor \( a \) from \( x^2 \) and \( x \) first.
  • Solving: once \((x+p)^2=q\), write \( x=-p\pm\sqrt{q} \).
  • Graphing: \( a(x+p)^2+q \) gives vertex \((-p,q)\) and axis \(x=-p\).
  • Keep one equals per line for method marks.

⚠️ Ways it Can Appear in an Exam

  • Rewrite \( ax^2+bx+c \) in completed-square form.
  • Show that \( ax^2+bx+c \) can be written as \( a(x+p)^2+q \).
  • Solve a quadratic by completing the square.
  • Find vertex and axis by completing the square.
  • Choose a parameter so an equation has one, two, or no real roots.
  • Match equations to sketches by comparing \( a,p,q \).
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Build speed and accuracy on completing the square in both the solving and graphing forms.

Next: The Quadratic Formula

Completing the Square: Practice Rooms

Randomly generated, auto-marked practice in completing the square for Edexcel IGCSE Maths. Rooms 1 to 3 ask for the completed-square form \(a(x+p)^2+q\) (with a = 1, then a not equal to 1, then a fractional value of p); Room 4 asks you to solve a quadratic using that form. Every grid gives 16 fresh questions. Type a square with the x² button or by typing ^2; as you type, your answer is shown in the box as real maths with a raised index. Each correct answer adds to your global streak.

✓ Correct 0
✗ Re-attempts 0
🔥 Streak 0
🏆 Best 0

Rooms 1 to 3: write the completed-square form, for example (x+3)^2-1 or 2(x+2)^2-2. Room 4: give both solutions separated by a comma, for example -1,-5. Order does not matter.