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Completing the Square0%

Functions & Graphs · Topic 6 of 9

Completing the Square

Video coming soon3 worked examples

Theory

Completing the square is a method used to write a quadratic expression ax2+bx+cax^2 + bx + c in the form a(x+p)2+qa(x + p)^2 + q.

This form is very useful for finding the turning point (vertex) of a parabola and solving quadratic equations.

The Process (when a = 1):

  1. Start with x2+bx+cx^2 + bx + c
  2. Halve the coefficient of xx (which is b/2b/2)
  3. Put it inside a squared bracket with xx: (x+b/2)2(x + b/2)^2
  4. Subtract the square of this number outside the bracket.
  5. Add the original constant cc.
  6. Simplify the constants.

The Turning Point:

For a quadratic in the form y=a(x+p)2+qy = a(x + p)^2 + q:

  • The turning point is at (−p,q)(-p, q).
  • If a>0a > 0, it is a minimum turning point.
  • If a<0a < 0, it is a maximum turning point.
  • The axis of symmetry is the line x=−px = -p.

⚠️ Common Examiner Traps

  • Halve the coefficient of xx, then square it: both steps, in that order. Halving without squaring, or squaring without halving, is the usual slip.
  • Subtract what you added: x2+6xx^2+6x becomes (x+3)2−9(x+3)^2 - 9. The −9-9 keeps the expression equal to what you started with.
  • Factor out a coefficient first: if the x2x^2 term has a coefficient, take it outside the bracket before completing the square, and remember it multiplies the correction term.
  • Read the turning point correctly: (x+3)2−9(x+3)^2 - 9 has its minimum at (−3,−9)(-3,-9) — the xx value flips sign, the yy value does not.
  • Watch signs with a negative coefficient: the parabola opens downwards, so the turning point is a maximum.

Worked examples

Example 1

Express x2+6x−2x^2 + 6x - 2 in the form (x+p)2+q(x + p)^2 + q.

Halve the coefficient of xx: 6÷2=36 \div 2 = 3.

=(x+3)2−(3)2−2=(x+3)2−9−2=(x+3)2−11\begin{aligned} &= (x + 3)^2 - (3)^2 - 2 \\ &= (x + 3)^2 - 9 - 2 \\ &= (x + 3)^2 - 11 \end{aligned}

Here p=3p = 3 and q=−11q = -11.

Example 2

Express 2x2−12x+72x^2 - 12x + 7 in the form a(x+p)2+qa(x + p)^2 + q.

Take out the common factor of 22 from the xx terms first.

=2(x2−6x)+7= 2(x^2 - 6x) + 7

Now complete the square on the bracket (x2−6x)(x^2 - 6x). Half of −6-6 is −3-3.

=2[(x−3)2−(−3)2]+7=2[(x−3)2−9]+7\begin{aligned} &= 2\left[(x - 3)^2 - (-3)^2\right] + 7 \\ &= 2\left[(x - 3)^2 - 9\right] + 7 \end{aligned}

Expand the square bracket by multiplying the −9-9 by the 22 outside:

=2(x−3)2−18+7=2(x−3)2−11\begin{aligned} &= 2(x - 3)^2 - 18 + 7 \\ &= 2(x - 3)^2 - 11 \end{aligned}

Example 3

Express 5−4x−x25 - 4x - x^2 in the form q−(x+p)2q - (x + p)^2.

Rewrite the expression in standard order:

=−x2−4x+5= -x^2 - 4x + 5

Factor out −1-1 from the xx terms:

=−(x2+4x)+5= -(x^2 + 4x) + 5

Complete the square inside the bracket:

=−[(x+2)2−(2)2]+5=−[(x+2)2−4]+5\begin{aligned} &= -\left[(x + 2)^2 - (2)^2\right] + 5 \\ &= -\left[(x + 2)^2 - 4\right] + 5 \end{aligned}

Expand by distributing the negative sign:

=−(x+2)2+4+5=9−(x+2)2\begin{aligned} &= -(x + 2)^2 + 4 + 5 \\ &= 9 - (x + 2)^2 \end{aligned}