Welcome to Completing the Square
Welcome to one of the most powerful tools in Unit 1: Pure Mathematics! Don't worry if this topic sounds a bit intimidating at first. By the end of these notes, you will see that completing the square is simply a reliable step-by-step puzzle that unlocks quadratic equations, graph sketches, and turning points.
Why is this useful?
Think of an ordinary quadratic expression like \(x^2 + 6x + 5\) as an unpacked suitcase. It has all its parts spread out. Completing the square is like neatly repacking that suitcase into the form \((x + p)^2 + q\). Once it is packed neatly, you can instantly see where a graph turns, find exact solutions without the quadratic formula, and prove properties about equations!
Did you know? The term "completing the square" comes from ancient geometry! Babylonian and Greek mathematicians literally drew squares and rectangles with tiles, cutting them up and adding a tiny square corner to make a complete larger square.
Section 1: The Basics (When \(a = 1\))
Let us begin with standard quadratic expressions where the number in front of \(x^2\) is just \(1\). These have the form \(x^2 + bx + c\).
The Big Idea: Perfect Squares
When you expand \((x + d)^2\), you always get:
\((x + d)^2 = x^2 + 2dx + d^2\)
Notice that the middle coefficient is \(2d\), and the end number is \(d^2\). So, to work backwards, we simply halve the middle number and then subtract its square.
The 3-Step Golden Rule
To rewrite \(x^2 + bx + c\) into the form \((x + p)^2 + q\):
Step 1: Halve the coefficient of \(x\). This gives \(p = \frac{b}{2}\). Put this inside the bracket: \(\left(x + \frac{b}{2}\right)^2\).
Step 2: Subtract the square of that new number: \(-\left(\frac{b}{2}\right)^2\).
Step 3: Add the original constant number \(c\) at the end and simplify.
The universal algebraic rule is:
\(x^2 + bx + c = \left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\)
Worked Example 1: Standard Positive Numbers
Write \(x^2 + 8x + 3\) in the form \((x + p)^2 + q\).
Step 1: Halve the middle number (\(+8\)) to get \(+4\):
\((x + 4)^2\)
Step 2: Subtract the square of \(4\) (which is \(4^2 = 16\)):
\((x + 4)^2 - 16\)
Step 3: Bring down the constant \(+3\) and simplify:
\((x + 4)^2 - 16 + 3 = (x + 4)^2 - 13\)
Final Answer: \((x + 4)^2 - 13\)
Worked Example 2: Negative Middle Term and Odd Numbers
Write \(x^2 - 5x + 2\) in the form \((x + p)^2 + q\).
Step 1: Halve the middle number (\(-5\)) to get \(-\frac{5}{2}\) or \(-2.5\):
\(\left(x - \frac{5}{2}\right)^2\)
Step 2: Subtract the square of \(\frac{5}{2}\), which is \(\left(\frac{5}{2}\right)^2 = \frac{25}{4}\):
\(\left(x - \frac{5}{2}\right)^2 - \frac{25}{4}\)
Step 3: Add the original constant \(+2\) (note that \(2 = \frac{8}{4}\)):
\(\left(x - \frac{5}{2}\right)^2 - \frac{25}{4} + \frac{8}{4} = \left(x - \frac{5}{2}\right)^2 - \frac{17}{4}\)
Final Answer: \(\left(x - \frac{5}{2}\right)^2 - \frac{17}{4}\)
Common Mistakes to Avoid
• Forgetting to subtract: Always subtract \(\left(\frac{b}{2}\right)^2\), even if \(b\) is negative. The term you subtract outside is always negative because squaring any real number produces a positive value.
• Halving \(x\) instead of the number: Write \((x + 3)^2\), never \(\left(\frac{x}{2} + 3\right)^2\).
Key Takeaways for Section 1
Memory Trick: "Halve it, square it, take it away!"
For any \(x^2 + bx\), replace it with \(\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2\).
Section 2: Harder Quadratics (When \(a \neq 1\))
In CCEA GCSE Further Mathematics, you will frequently meet quadratics where the coefficient of \(x^2\) is not \(1\), such as \(2x^2 + 12x - 5\) or \(-x^2 + 6x + 1\). The target form is usually \(a(x + p)^2 + q\).
Step-by-Step Method for \(ax^2 + bx + c\)
Step 1: Factor out the coefficient \(a\) from the terms containing \(x^2\) and \(x\). (You can leave the constant \(c\) outside the brackets or include it inside; keeping it outside is usually safer!).
Step 2: Complete the square inside the bracket using the standard rule.
Step 3: Expand the outer bracket by multiplying through by \(a\).
Step 4: Combine the constant numbers at the end.
Worked Example 3: When \(a > 1\)
Express \(2x^2 + 12x + 7\) in the form \(a(x + p)^2 + q\).
Step 1: Factor out \(2\) from the first two terms:
\(2[x^2 + 6x] + 7\)
Step 2: Complete the square inside the square brackets:
Halve \(6\) to get \(3\), then subtract \(3^2 = 9\):
\(2\left[(x + 3)^2 - 9\right] + 7\)
Step 3: Multiply the \(2\) back across the terms inside:
\(2(x + 3)^2 - 18 + 7\)
Step 4: Simplify the constant numbers:
\(2(x + 3)^2 - 11\)
Final Answer: \(a = 2\), \(p = 3\), \(q = -11\)
Worked Example 4: When \(a\) is Negative
Express \(-x^2 + 8x - 5\) in the form \(q - (x + p)^2\) or \(a(x + p)^2 + q\).
Step 1: Factor out \(-1\) from the \(x\) terms:
\(-[x^2 - 8x] - 5\)
Step 2: Complete the square inside the brackets:
\(-[(x - 4)^2 - 16] - 5\)
Step 3: Multiply through by the negative sign:
\(-(x - 4)^2 + 16 - 5\)
Step 4: Simplify:
\(-(x - 4)^2 + 11\) or \(11 - (x - 4)^2\)
Key Takeaways for Section 2
Whenever \(a \neq 1\), factor out \(a\) first before touching the middle term. Remember to multiply \(a\) back by the subtracted square value before adding the constant!
Section 3: Applications of Completing the Square
Completing the square is not just an algebraic exercise—it is a master key that solves three major problems in Further Mathematics.
Application 1: Finding Turning Points and Lines of Symmetry
For any quadratic curve \(y = a(x + p)^2 + q\):
• Because a squared number \((x + p)^2\) is always greater than or equal to zero (\(\ge 0\)), its smallest possible value is \(0\), which happens when \(x + p = 0 \implies x = -p\).
• Minimum Point: If \(a > 0\) (a U-shaped curve \(\cup\)), the minimum turning point is \((-p, q)\).
• Maximum Point: If \(a < 0\) (an inverted curve \(\cap\)), the maximum turning point is \((-p, q)\).
• Line of Symmetry: The vertical mirror line running through the vertex is always \(x = -p\).
Quick Review Box:
If \(y = 3(x - 4)^2 + 10\):
• Turning point = \((4, 10)\) [Minimum because \(a = +3\)]
• Minimum value of \(y\) = \(10\)
• Equation of the line of symmetry: \(x = 4\)
Application 2: Solving Quadratic Equations (Exact Values)
You can solve equations without using the quadratic formula by rearranging the completed square.
Worked Example 5: Solve \(x^2 - 6x - 2 = 0\), giving your answer in exact surd form.
Step 1: Complete the square:
\((x - 3)^2 - 9 - 2 = 0\)
\((x - 3)^2 - 11 = 0\)
Step 2: Isolate the bracket:
\((x - 3)^2 = 11\)
Step 3: Take the square root of both sides (remember \(\pm\)):
\(x - 3 = \pm\sqrt{11}\)
Step 4: Solve for \(x\):
\(x = 3 \pm \sqrt{11}\)
Application 3: Algebraic Proofs (Always Positive / Negative)
Exam questions may ask: "Prove that \(x^2 + 4x + 7 > 0\) for all real values of \(x\)."
Proof:
1. Complete the square: \(x^2 + 4x + 7 = (x + 2)^2 - 4 + 7 = (x + 2)^2 + 3\).
2. State the property: Since \((x + 2)^2 \ge 0\) for all real \(x\), the minimum value of the expression is \(0 + 3 = 3\).
3. Conclude: Since \(3 > 0\), \((x + 2)^2 + 3 > 0\) for all values of \(x\).
Quick Reference Summary
1. Form \(x^2 + bx + c\):
\(\left(x + \frac{b}{2}\right)^2 - \left(\frac{b}{2}\right)^2 + c\)
2. Form \(ax^2 + bx + c\):
Factor out \(a\), complete the square inside, and multiply back through.
3. Coordinates of the Vertex:
For \(y = a(x + p)^2 + q\), the turning point is \((-p, q)\). Note the sign flip for the x-coordinate!
4. Symmetry:
The line of symmetry is always the vertical line \(x = -p\).