Completing the square turns quadratic expressions into a more useful form. It’s a way of rewriting a quadratic expression like \(ax^2 + bx + c\) into the form \(a(x + p)^2 + q\). While it might seem pointless, it’s actually very useful. It’s makes sketching their graphs, and finding the minimum or maximum values really easy. It also gives us another way to solving them.
Everyone finds completing the square tricky at first, so don’t worry when you do. Just give it time, and with a bit of practice and understanding the pattern, it becomes quite straightforward. It actually connects the algebra of the equation to the geometry and shape of the graph. If you go on to do A-Level maths you’ll use it a lot.
TLDR
- Completing the square transforms a quadratic expression \(x^2 + bx + c\) into the form \((x + p)^2 + q\)
- For expressions of the form \(x^2 + bx + c\):
- Find \(p\) by taking half of the \(x\) coefficient: \(p = \dfrac{b}{2}\)
- Form \((x + p)^2 = x^2 + 2px + p^2 = x^2 + bx + \dfrac{b^2}{4}\)
- Find \(q\) by subtracting \(\dfrac{b^2}{4}\) from \(c\): \(q = c – \dfrac{b^2}{4}\)
- Final form: \((x + \dfrac{b}{2})^2 + (c – \dfrac{b^2}{4})\)
- For expressions of the form \(ax^2 + bx + c\) where \(a \neq 1\):
- Factor out \(a\) first: \(a(x^2 + \dfrac{b}{a}x + \dfrac{c}{a})\)
- Complete the square for the expression inside the brackets
- The final form looks like: \(a(x + \dfrac{b}{2a})^2 + (c – \dfrac{b^2}{4a})\)
- Applications include solving quadratic equations, finding the vertex of a parabola, and determining minimum/maximum values
What You Need to Know
- Expand products of two or more binomials and factorise quadratic expressions of the form \(ax^2 + bx + c\)
- Identify and interpret roots, intercepts, and turning points of quadratic functions graphically; deduce roots algebraically and deduce turning points by completing the square
- Solve quadratic equations by completing the square and using the quadratic formula
Key Facts
🔑 Key Fact: For any quadratic expression \(x^2 + bx + c\), the completed square form is \((x + \dfrac{b}{2})^2 + (c – \dfrac{b^2}{4})\).
🔑 Key Fact: The minimum or maximum value of a quadratic function occurs at \(x = -\dfrac{b}{2a}\), which you can find directly from the completed square form \((x + p)^2 + q\).
🔑 Key Fact: When the coefficient of \(x^2\) is not 1, always factor it out first before completing the square.
🔑 Key Fact: The completed square form \((x + p)^2 + q\) gives you the top or bottom of the curve as the point \((-p, q)\)
Completing the Square: Basic Form
Let’s start with the simplest case: completing the square for expressions of the form \(x^2 + bx + c\).
The Pattern
The key insight for completing the square comes from the expansion of \((x + p)^2\):
\((x + p)^2 = x^2 + 2px + p^2\)
Looking at this expansion, if we have \(x^2 + bx\), we need to find the value of \(p\) such that \(2p = b\), which means \(p = \frac{b}{2}\).
Example 1
Complete the square for \(x^2 + 6x + 8\).
Step 1: Identify the coefficient of \(x\), which is 6, and calculate \(p = \dfrac{b}{2} = \dfrac{6}{2} = 3\).
Step 2: Form \((x + p)^2 = (x + 3)^2 = x^2 + 6x + 9\).
Step 3: Compare with our original expression:
\(x^2 + 6x + 8\)
\(x^2 + 6x + 9 – 1\) (We’ve added and subtracted 1 to maintain equality)
Step 4: Rewrite as:
\((x + 3)^2 – 1\)
Therefore, \(x^2 + 6x + 8 = (x + 3)^2 – 1\).
Example 2
Complete the square for \(x^2 – 8x + 7\).
Step 1: Identify the coefficient of \(x\), which is -8, and calculate \(p = \dfrac{b}{2} = \dfrac{-8}{2} = -4\).
Step 2: Form \((x + p)^2 = (x – 4)^2 = x^2 – 8x + 16\).
Step 3: Compare with our original expression:
\(x^2 – 8x + 7\)
\(x^2 – 8x + 16 – 9\) (We’ve added and subtracted 9 to maintain equality)
Step 4: Rewrite as:
\((x – 4)^2 – 9\)
Therefore, \(x^2 – 8x + 7 = (x – 4)^2 – 9\).
Direct Method for Basic Form
There’s a quicker way to complete the square using a formula.
For \(x^2 + bx + c\), the completed square form is:
\((x + \dfrac{b}{2})^2 + (c – \dfrac{b^2}{4})\)
Example 3
Complete the square for \(x^2 + 10x + 23\).
Step 1: Identify \(b = 10\) and \(c = 23\).
Step 2: Calculate \(\dfrac{b}{2} = \dfrac{10}{2} = 5\).
Step 3: Calculate \(\dfrac{b^2}{4} = \dfrac{10^2}{4} = \dfrac{100}{4} = 25\).
Step 4: Calculate \(c – \dfrac{b^2}{4} = 23 – 25 = -2\).
Step 5: Write the completed square form:
\((x + 5)^2 – 2\)
Therefore, \(x^2 + 10x + 23 = (x + 5)^2 – 2\).
Let’s verify this is correct by expanding \((x + 5)^2 – 2\):
\((x + 5)^2 – 2 = x^2 + 10x + 25 – 2 = x^2 + 10x + 23\) ✓
Completing the Square: General Form
Now let’s look at the general case where the coefficient of \(x^2\) is not 1: \(ax^2 + bx + c\).
The Method
For \(ax^2 + bx + c\), where \(a \neq 1\):
- Factor out \(a\) from the terms with \(x\): \(a(x^2 + \frac{b}{a}x) + c\)
- Complete the square for the expression inside the brackets
- Rearrange the result to the final form
Example 4
Complete the square for \(2x^2 + 12x + 13\).
Step 1: Factor out 2 from the terms with \(x\).
\(2x^2 + 12x + 13 = 2(x^2 + 6x) + 13\)
Step 2: Complete the square for \(x^2 + 6x\).
For \(x^2 + 6x\), we have \(b = 6\), so \(p = \dfrac{b}{2} = 3\).
\(x^2 + 6x = (x + 3)^2 – 9\)
Step 3: Substitute this back.
\(2(x^2 + 6x) + 13 = 2((x + 3)^2 – 9) + 13\)
\(= 2(x + 3)^2 – 18 + 13\)
\(= 2(x + 3)^2 – 5\)
Therefore, \(2x^2 + 12x + 13 = 2(x + 3)^2 – 5\).
Example 5
Complete the square for \(3x^2 – 6x + 5\).
Step 1: Factor out 3 from the terms with \(x\).
\(3x^2 – 6x + 5 = 3(x^2 – 2x) + 5\)
Step 2: Complete the square for \(x^2 – 2x\).
For \(x^2 – 2x\), we have \(b = -2\), so \(p = \dfrac{b}{2} = -1\).
\(x^2 – 2x = (x – 1)^2 – 1\)
Step 3: Substitute this back.
\(3(x^2 – 2x) + 5 = 3((x – 1)^2 – 1) + 5\)
\(= 3(x – 1)^2 – 3 + 5\)
\(= 3(x – 1)^2 + 2\)
Therefore, \(3x^2 – 6x + 5 = 3(x – 1)^2 + 2\).
Direct Method for General Form
For \(ax^2 + bx + c\), the completed square form is:
\(a(x + \dfrac{b}{2a})^2 + (c – \dfrac{b^2}{4a})\)
Example 6
Complete the square for \(4x^2 – 16x + 25\).
Step 1: Identify \(a = 4\), \(b = -16\), and \(c = 25\).
Step 2: Calculate \(\dfrac{b}{2a} = \dfrac{-16}{2 \times 4} = \dfrac{-16}{8} = -2\).
Step 3: Calculate \(\dfrac{b^2}{4a} = \dfrac{(-16)^2}{4 \times 4} = \dfrac{256}{16} = 16\).
Step 4: Calculate \(c – \dfrac{b^2}{4a} = 25 – 16 = 9\).
Step 5: Write the completed square form:
\(4(x – 2)^2 + 9\)
Therefore, \(4x^2 – 16x + 25 = 4(x – 2)^2 + 9\).
Let’s verify this is correct by expanding \(4(x – 2)^2 + 9\):
\(4(x – 2)^2 + 9 = 4(x^2 – 4x + 4) + 9 = 4x^2 – 16x + 16 + 9 = 4x^2 – 16x + 25\) ✓
Applications of Completing the Square
1. Solving Quadratic Equations
When a quadratic equation is written in the form \((x + p)^2 = d\), we can solve it easily by taking the square root of both sides.
Example 7
Solve the equation \(x^2 + 6x + 8 = 0\) by completing the square.
Step 1: Move the constant term to the right-hand side.
\(x^2 + 6x = -8\)
Step 2: Complete the square on the left.
\(x^2 + 6x + 9 = -8 + 9\)
\((x + 3)^2 = 1\)
Step 3: Take the square root of both sides.
\(x + 3 = \pm 1\)
\(x = -3 \pm 1\)
\(x = -2\) or \(x = -4\)
Therefore, the solutions are \(x = -2\) and \(x = -4\).
2. Finding the Coordinates of the Maximum or Minimum
The point at the top or bottom of a quadratic graph is known as the maximum or minimum. It’s also known as the vertex of the parabola. If we have the quadratic equation \(y = ax^2 + bx + c\) the vertex can be found directly from the completed square form.
If \(y = a(x + p)^2 + q\), the vertex is at \((-p, q)\).
Example 8
Find the coordinates of the vertex of the parabola \(y = 2x^2 + 12x + 13\).
From Example 4, we know that \(2x^2 + 12x + 13 = 2(x + 3)^2 – 5\).
In the form \(a(x + p)^2 + q\), we have \(p = 3\) and \(q = -5\).
Therefore, the vertex is at \((-p, q) = (-3, -5)\).
3. Finding Minimum or Maximum Values
For a quadratic function \(y = ax^2 + bx + c\):
- If \(a > 0\), the function has a minimum value
- If \(a < 0\), the function has a maximum value
This minimum or maximum occurs at \(x = -\frac{b}{2a}\) and the value is \(c – \frac{b^2}{4a}\).
Note that this is just the y-value of the vertex we were calculating above.
Example 9
Find the minimum value of \(f(x) = 3x^2 – 6x + 5\) and the value of \(x\) where this occurs.
From Example 5, we know that \(3x^2 – 6x + 5 = 3(x – 1)^2 + 2\).
Since \(a = 3 > 0\), the function has a minimum value.
The minimum occurs at \(x = 1\) (from \(x – 1 = 0\)) and the minimum value is \(2\).
Common Mistakes to Avoid
- Forgetting to factor out the coefficient of \(x^2\) when it’s not 1
- Incorrect: Trying to complete the square for \(2x^2 + 12x + 13\) as \((x + 6)^2 + k\)
- Correct: Factor out 2 first: \(2(x^2 + 6x) + 13\) and then complete the square inside the brackets
- Miscalculating the constant term
- Incorrect: \(x^2 + 6x + 8 = (x + 3)^2 + 8\) (forgetting to subtract \(p^2\))
- Correct: \(x^2 + 6x + 8 = (x + 3)^2 – 9 + 8 = (x + 3)^2 – 1\)
- Handling negative coefficients incorrectly
- Incorrect: \(x^2 – 4x + 7\) as \((x + 2)^2 + 3\) (should be \(x – 2\), not \(x + 2\))
- Correct: \(x^2 – 4x + 7 = (x – 2)^2 + 3\)
- Forgetting the final simplification step
- Incorrect: Leaving the answer as \(2(x + 3)^2 – 18 + 13\)
- Correct: Simplifying to \(2(x + 3)^2 – 5\)
- Using the wrong method to find the vertex
- Incorrect: Stating that the vertex of \(y = 2(x + 3)^2 – 5\) is at \((3, -5)\)
- Correct: The vertex is at \((-3, -5)\) because the form is \((x + p)^2\), not \((x – p)^2\)
Questions
Try these questions to practice completing the square:
- Complete the square for \(x^2 + 8x + 12\).
- Complete the square for \(x^2 – 10x + 20\).
- Complete the square for \(2x^2 + 16x + 30\).
- Complete the square for \(3x^2 – 12x + 7\).
- Find the vertex of the parabola \(y = x^2 + 6x – 5\).
- Find the minimum value of \(f(x) = 2x^2 – 8x + 9\).
- Solve the equation \(x^2 – 14x + 45 = 0\) by completing the square.
- Solve the equation \(3x^2 + 6x – 9 = 0\) by completing the square.
- Determine, by completing the square, whether the quadratic \(4x^2 – 4x + 3\) can be factorised.
- Find the range of values of \(k\) for which the equation \(2x^2 + 4x + k = 0\) has real solutions.
Solutions
Question 1
Complete the square for \(x^2 + 8x + 12\).
Step 1: Identify \(b = 8\), so \(p = \frac{b}{2} = 4\).
Step 2: Calculate \(p^2 = 4^2 = 16\).
Step 3: Rewrite the expression:
\(x^2 + 8x + 12 = x^2 + 8x + 16 – 16 + 12 = (x + 4)^2 – 4\)
Therefore, \(x^2 + 8x + 12 = (x + 4)^2 – 4\).
Question 2
Complete the square for \(x^2 – 10x + 20\).
Step 1: Identify \(b = -10\), so \(p = \frac{b}{2} = -5\).
Step 2: Calculate \(p^2 = (-5)^2 = 25\).
Step 3: Rewrite the expression:
\(x^2 – 10x + 20 = x^2 – 10x + 25 – 25 + 20 = (x – 5)^2 – 5\)
Therefore, \(x^2 – 10x + 20 = (x – 5)^2 – 5\).
Question 3
Complete the square for \(2x^2 + 16x + 30\).
Step 1: Factor out 2 from the terms with \(x\).
\(2x^2 + 16x + 30 = 2(x^2 + 8x) + 30\)
Step 2: Complete the square for \(x^2 + 8x\).
For \(x^2 + 8x\), we have \(b = 8\), so \(p = \frac{b}{2} = 4\).
\(x^2 + 8x = (x + 4)^2 – 16\)
Step 3: Substitute this back.
\(2(x^2 + 8x) + 30 = 2((x + 4)^2 – 16) + 30\)
\(= 2(x + 4)^2 – 32 + 30\)
\(= 2(x + 4)^2 – 2\)
Therefore, \(2x^2 + 16x + 30 = 2(x + 4)^2 – 2\).
Question 4
Complete the square for \(3x^2 – 12x + 7\).
Step 1: Factor out 3 from the terms with \(x\).
\(3x^2 – 12x + 7 = 3(x^2 – 4x) + 7\)
Step 2: Complete the square for \(x^2 – 4x\).
For \(x^2 – 4x\), we have \(b = -4\), so \(p = \frac{b}{2} = -2\).
\(x^2 – 4x = (x – 2)^2 – 4\)
Step 3: Substitute this back.
\(3(x^2 – 4x) + 7 = 3((x – 2)^2 – 4) + 7\)
\(= 3(x – 2)^2 – 12 + 7\)
\(= 3(x – 2)^2 – 5\)
Therefore, \(3x^2 – 12x + 7 = 3(x – 2)^2 – 5\).
Question 5
Find the vertex of the parabola \(y = x^2 + 6x – 5\).
Step 1: Complete the square for \(x^2 + 6x\).
For \(x^2 + 6x\), we have \(b = 6\), so \(p = \frac{b}{2} = 3\).
\(x^2 + 6x = (x + 3)^2 – 9\)
Step 2: Rewrite the equation.
\(y = x^2 + 6x – 5 = (x + 3)^2 – 9 – 5 = (x + 3)^2 – 14\)
Step 3: In the form \(y = (x + p)^2 + q\), we have \(p = 3\) and \(q = -14\).
Therefore, the vertex is at \((-p, q) = (-3, -14)\).
Question 6
Find the minimum value of \(f(x) = 2x^2 – 8x + 9\).
Step 1: Factor out 2 from the terms with \(x\).
\(f(x) = 2x^2 – 8x + 9 = 2(x^2 – 4x) + 9\)
Step 2: Complete the square for \(x^2 – 4x\).
For \(x^2 – 4x\), we have \(b = -4\), so \(p = \frac{b}{2} = -2\).
\(x^2 – 4x = (x – 2)^2 – 4\)
Step 3: Substitute this back.
\(f(x) = 2(x^2 – 4x) + 9 = 2((x – 2)^2 – 4) + 9\)
\(= 2(x – 2)^2 – 8 + 9\)
\(= 2(x – 2)^2 + 1\)
Step 4: Since \(a = 2 > 0\), the function has a minimum value.
The minimum occurs at \(x = 2\) (from \(x – 2 = 0\)) and the minimum value is \(1\).
Question 7
Solve the equation \(x^2 – 14x + 45 = 0\) by completing the square.
Step 1: Move the constant term to the right-hand side.
\(x^2 – 14x = -45\)
Step 2: Complete the square on the left.
\(x^2 – 14x + 49 = -45 + 49\)
\((x – 7)^2 = 4\)
Step 3: Take the square root of both sides.
\(x – 7 = \pm 2\)
\(x = 7 \pm 2\)
\(x = 9\) or \(x = 5\)
Therefore, the solutions are \(x = 9\) and \(x = 5\).
Question 8
Solve the equation \(3x^2 + 6x – 9 = 0\) by completing the square.
Step 1: Factor out 3.
\(3(x^2 + 2x – 3) = 0\)
Step 2: Since 3 ≠ 0, we need to solve \(x^2 + 2x – 3 = 0\).
Step 3: Move the constant term to the right-hand side.
\(x^2 + 2x = 3\)
Step 4: Complete the square on the left.
\(x^2 + 2x + 1 = 3 + 1\)
\((x + 1)^2 = 4\)
Step 5: Take the square root of both sides.
\(x + 1 = \pm 2\)
\(x = -1 \pm 2\)
\(x = 1\) or \(x = -3\)
Therefore, the solutions are \(x = 1\) and \(x = -3\).
Question 9
Determine, by completing the square, whether the quadratic \(4x^2 – 4x + 3\) can be factorised.
Step 1: Factor out 4.
\(4x^2 – 4x + 3 = 4(x^2 – x) + 3\)
Step 2: Complete the square for \(x^2 – x\).
For \(x^2 – x\), we have \(b = -1\), so \(p = \frac{b}{2} = -\frac{1}{2}\).
\(x^2 – x = (x – \frac{1}{2})^2 – \frac{1}{4}\)
Step 3: Substitute this back.
\(4(x^2 – x) + 3 = 4((x – \frac{1}{2})^2 – \frac{1}{4}) + 3\)
\(= 4(x – \frac{1}{2})^2 – 4 \cdot \frac{1}{4} + 3\)
\(= 4(x – \frac{1}{2})^2 – 1 + 3\)
\(= 4(x – \frac{1}{2})^2 + 2\)
Step 4: For a quadratic to be factorisable over the real numbers, it needs to be expressible as \((px + q)(rx + s)\) where \(p, q, r, s\) are real numbers. This is only possible if the constant term after completing the square is less than or equal to 0.
Since the constant term is \(+2 > 0\), the quadratic \(4x^2 – 4x + 3\) cannot be factorised over the real numbers.
Question 10
Find the range of values of \(k\) for which the equation \(2x^2 + 4x + k = 0\) has real solutions.
Step 1: Factor out 2.
\(2x^2 + 4x + k = 2(x^2 + 2x) + k\)
Step 2: Complete the square for \(x^2 + 2x\).
For \(x^2 + 2x\), we have \(b = 2\), so \(p = \frac{b}{2} = 1\).
\(x^2 + 2x = (x + 1)^2 – 1\)
Step 3: Substitute this back.
\(2(x^2 + 2x) + k = 2((x + 1)^2 – 1) + k\)
\(= 2(x + 1)^2 – 2 + k\)
Step 4: For the equation to have real solutions, we need the discriminant to be greater than or equal to 0. In the form \(a(x + p)^2 + q = 0\), the equation has real solutions if \(q \leq 0\).
In our case, \(q = -2 + k\), so we need \(-2 + k \leq 0\), which gives \(k \leq 2\).
Therefore, the equation \(2x^2 + 4x + k = 0\) has real solutions when \(k \leq 2\).
Summary
- Completing the square is a technique to rewrite a quadratic expression into the form \(a(x + p)^2 + q\)
- For expressions where the coefficient of \(x^2\) is 1, use \((x + \frac{b}{2})^2 + (c – \frac{b^2}{4})\)
- For expressions where the coefficient of \(x^2\) is not 1, factor it out first, then complete the square
- The vertex of a parabola \(y = a(x + p)^2 + q\) is at the point \((-p, q)\)
- A quadratic function has a minimum value when \(a > 0\) or a maximum value when \(a < 0\)
- Completing the square helps solve quadratic equations by taking square roots
- The technique provides a direct link between algebraic expressions and geometric properties of quadratic functions
Understanding completing the square as well as allowing you to answer questions on them also build your algebra skills, and develops your familiarity with graphs. It is a challenging topic so give yourself time to get familiar with it. Good luck!
By Atomic