Home / A-Level Maths 9709 / Quadratics / Completing the square
Cambridge International AS & A Level · Topic 1.1
A-Level Maths 9709 / Topic 1.1 / AS & A Level

Every parabola has a vertex.

Rewrite any quadratic as a square plus a constant and its graph gives up its secrets: the vertex, the line of symmetry, the least or greatest value and the range, all without a single derivative.

Full-screen labAlgebra lab → Topic 1.1All quadratics lessons →
The key idea

\(x^{2} + bx + c = \left(x + \dfrac{b}{2}\right)^{2} - \left(\dfrac{b}{2}\right)^{2} + c\). In the form \(a(x + p)^{2} + q\), the vertex is \((-p,\ q)\): a minimum if \(a > 0\), a maximum if \(a < 0\).

x y 1 3 5 5 vertex (3, −4) x = 3
Fig. 1: y = x² - 6x + 5 = (x - 3)² - 4 has its vertex at (3, -4), line of symmetry x = 3, roots 1 and 5 and y-intercept 5
Section 01

Half the coefficient, squared.

In the lab, press Make the square and watch \(x^{2} + bx\) rearranged into a square: there is always a small corner missing, of area \(\left(\frac{b}{2}\right)^{2}\). Completing the square means adding that corner and then taking it away again so the value does not change.

Worked example Integer half
Express \(x^{2} - 6x + 11\) in the form \((x - a)^{2} + b\).
  1. Half of \(-6\) is \(-3\), so start with \((x - 3)^{2} = x^{2} - 6x + 9\).
  2. That bracket contains an extra \(9\) we did not have, so subtract it: \(x^{2} - 6x = (x - 3)^{2} - 9\).
  3. Add the constant: \((x - 3)^{2} - 9 + 11 = (x - 3)^{2} + 2\).
Answer: \((x - 3)^{2} + 2\), so \(a = 3,\ b = 2\)
Worked example Fractional half
Express \(x^{2} + 5x + 1\) in the form \((x + p)^{2} + q\).
  1. Half of \(5\) is \(\dfrac{5}{2}\), and \(\left(\dfrac{5}{2}\right)^{2} = \dfrac{25}{4}\).
  2. \(x^{2} + 5x + 1 = \left(x + \dfrac{5}{2}\right)^{2} - \dfrac{25}{4} + 1\).
  3. \(-\dfrac{25}{4} + \dfrac{4}{4} = -\dfrac{21}{4}\).
Answer: \(\left(x + \dfrac{5}{2}\right)^{2} - \dfrac{21}{4}\)
Your turn
Express \(x^{2} + 8x + 3\) in the form \((x + a)^{2} + b\).
Examiner trap

The sign inside the bracket matches the sign of \(b\): \(x^{2} - 6x\) gives \((x - 3)^{2}\), not \((x + 3)^{2}\). And the square you subtract is always positive: \(-(-3)^{2} = -9\).

Section 02

When \(a\) is not 1.

For \(ax^{2} + bx + c\), take the factor \(a\) out of the \(x^{2}\) and \(x\) terms only, complete the square inside the bracket, then multiply back out carefully. A negative \(a\) works the same way and gives a maximum.

Worked example Positive a
Express \(2x^{2} - 12x + 7\) in the form \(a(x + b)^{2} + c\).
  1. Factor 2 from the first two terms: \(2(x^{2} - 6x) + 7\).
  2. Complete the square inside: \(x^{2} - 6x = (x - 3)^{2} - 9\).
  3. So \(2\big[(x - 3)^{2} - 9\big] + 7 = 2(x - 3)^{2} - 18 + 7\). The \(-9\) is multiplied by 2.
  4. \(-18 + 7 = -11\).
Answer: \(2(x - 3)^{2} - 11\): \(a = 2,\ b = -3,\ c = -11\)
Worked example Negative a
Express \(5 + 4x - x^{2}\) in the form \(a - (x - b)^{2}\).
  1. Take out \(-1\) from the \(x\) terms: \(5 - (x^{2} - 4x)\).
  2. \(x^{2} - 4x = (x - 2)^{2} - 4\).
  3. \(5 - \big[(x - 2)^{2} - 4\big] = 5 + 4 - (x - 2)^{2}\).
Answer: \(9 - (x - 2)^{2}\), so \(a = 9,\ b = 2\)
Examiner trap

Writing \(2x^{2} - 12x + 7 = 2(x - 3)^{2} - 9 + 7\) is the classic slip: the 9 sits INSIDE the bracket that is multiplied by 2, so it must become 18. Expand your answer to catch it.

Stage 1 · Learn

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Four quick checks. Each correct answer earns XP and records this skill on your profile.

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Express \(x^{2} + 10x + 30\) in the form \((x + a)^{2} + b\).

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What are the coordinates of the vertex of \(y = (x + 2)^{2} - 7\)?

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What is the least value of \(3(x - 1)^{2} + 4\)?

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Given \(2x^{2} + 8x + 1 = 2(x + 2)^{2} + q\), find \(q\).

Section 03

Reading the graph.

A square is never negative. That one fact turns the completed square form into the vertex, the least or greatest value, the line of symmetry and the range of the quadratic. It also solves equations without the formula.

FormVertexTypeRange (\(x \in \mathbb{R}\))
\((x - 3)^{2} - 4\)\((3,\ -4)\)minimum\(y \ge -4\)
\(2(x + 1)^{2} + 5\)\((-1,\ 5)\)minimum\(y \ge 5\)
\(9 - (x - 2)^{2}\)\((2,\ 9)\)maximum\(y \le 9\)
Worked example Vertex and sketch
For \(y = x^{2} - 6x + 5\), find the vertex and sketch the curve (Fig. 1).
  1. Complete the square: \(x^{2} - 6x + 5 = (x - 3)^{2} - 9 + 5 = (x - 3)^{2} - 4\).
  2. \((x - 3)^{2} \ge 0\), with equality only when \(x = 3\), so the least value of \(y\) is \(-4\).
  3. Roots: \((x - 3)^{2} = 4\), so \(x - 3 = \pm 2\), giving \(x = 1\) or \(x = 5\).
  4. The \(y\)-intercept is the constant term, \(5\).
Answer: Minimum point \((3,\ -4)\), symmetric about \(x = 3\), crossing the axes at \((1, 0)\), \((5, 0)\) and \((0, 5)\)
Worked example Solving with surds
Solve \(x^{2} - 6x + 4 = 0\) by completing the square, giving exact answers.
  1. \(x^{2} - 6x + 4 = (x - 3)^{2} - 9 + 4 = (x - 3)^{2} - 5\).
  2. \((x - 3)^{2} - 5 = 0 \Rightarrow (x - 3)^{2} = 5\).
  3. Square root both sides, keeping both signs: \(x - 3 = \pm\sqrt{5}\).
Answer: \(x = 3 + \sqrt{5}\) or \(x = 3 - \sqrt{5}\)
Examiner trap

The vertex of \(a(x + p)^{2} + q\) is at \(x = -p\), not \(x = p\): \((x + 1)^{2}\) is zero when \(x = -1\). Read the \(x\)-coordinate as the value that makes the bracket zero.

Stage 2 · Exam

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Express \(4x^{2} - 24x + 41\) in the form \(a(x + b)^{2} + c\).

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Find the greatest value of \(7 - 6x - x^{2}\) and the value of \(x\) at which it occurs.

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The function f is defined by \(f(x) = 2x^{2} - 8x + 3\) for \(x \in \mathbb{R}\). Find the range of f.

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The curve \(y = x^{2} + px + q\) has its vertex at \((4,\ -3)\). Find \(p\) and \(q\).

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Six original Cambridge-style questions
Completed square form with \(a = 1\), \(a \ne 1\) and \(a < 0\), vertex and range, solving with surds, a proof that a quadratic is always positive, and the height of a thrown ball.