The Central Idea
A quadratic expression such as
\[ x^2+6x+5 \]does not initially contain an obvious squared binomial.
But we can rewrite part of it:
\[ x^2+6x+9=(x+3)^2. \]Since the original expression has \(5\), not \(9\), we compensate by subtracting \(4\):
\[ x^2+6x+5 = x^2+6x+9-4. \]Therefore,
\[ x^2+6x+5 = (x+3)^2-4. \]The expression has changed form, but not value. This process is called completing the square.
Perfect Square Trinomials
Completing the square is based on two identities:
\[ (a+b)^2=a^2+2ab+b^2 \]and
\[ (a-b)^2=a^2-2ab+b^2. \]For example,
\[ (x+4)^2 = x^2+8x+16. \]Likewise,
\[ (x-5)^2 = x^2-10x+25. \]A quadratic trinomial that matches one of these patterns is called a perfect square trinomial.
The Missing Number
Suppose we begin with
\[ x^2+8x. \]We want to add a number that makes this into a perfect square trinomial.
Compare
\[ (x+a)^2=x^2+2ax+a^2. \]The middle coefficient is \(2a\).
Since
\[ 2a=8, \]we get
\[ a=4. \]So the needed constant is
\[ a^2=16. \]Therefore,
\[ x^2+8x+16=(x+4)^2. \]The Half-and-Square Rule
For an expression of the form
\[ x^2+bx, \]the number needed to complete the square is
\[ \left(\frac{b}{2}\right)^2. \]So
\[ x^2+bx+\left(\frac{b}{2}\right)^2 = \left(x+\frac{b}{2}\right)^2. \]This gives a reliable procedure:
- Take the coefficient of \(x\).
- Divide it by \(2\).
- Square the result.
That number completes the square.
Example: Positive Middle Term
Complete the square for
\[ x^2+10x. \]Half of \(10\) is
\[ 5. \]Square it:
\[ 5^2=25. \]Therefore,
\[ x^2+10x+25=(x+5)^2. \]Example: Negative Middle Term
Complete the square for
\[ x^2-12x. \]Half of \(-12\) is
\[ -6. \]Square it:
\[ (-6)^2=36. \]Therefore,
\[ x^2-12x+36=(x-6)^2. \]The sign inside the squared binomial comes from the half-value before it is squared.
Rewriting a Quadratic Expression
Consider
\[ x^2+8x+3. \]To create a perfect square, add and subtract \(16\):
\[ x^2+8x+16-16+3. \]Group the perfect square:
\[ (x+4)^2-13. \]Thus,
\[ x^2+8x+3=(x+4)^2-13. \]Adding and subtracting the same quantity preserves the value of the expression.
Why We Add and Subtract the Same Number
Suppose we change
\[ x^2+6x+2 \]into
\[ x^2+6x+9. \]That would change the value of the expression by \(7\).
To preserve equivalence, we must compensate:
\[ x^2+6x+2 = x^2+6x+9-7. \]Then
\[ x^2+6x+2 = (x+3)^2-7. \]Completing the square is therefore a controlled replacement: we create a useful pattern and then correct for what we added.
Completing the Square in an Equation
When solving an equation, there is another way to preserve equality: add the same quantity to both sides.
Consider
\[ x^2+6x=7. \]Half of \(6\) is \(3\), and
\[ 3^2=9. \]Add \(9\) to both sides:
\[ x^2+6x+9=16. \]The left side factors:
\[ (x+3)^2=16. \]Then
\[ x+3=\pm4. \]So
\[ x=1 \]or
\[ x=-7. \]Why the \(\pm\) Appears
From
\[ (x+3)^2=16, \]the quantity \(x+3\) could be either \(4\) or \(-4\), because both satisfy the equation:
\[ 4^2=16 \]and
\[ (-4)^2=16. \]So taking square roots while solving an equation gives
\[ x+3=\pm4. \]This differs from evaluating the principal square root
\[ \sqrt{16}=4. \]The equation has two possible signed values whose squares equal \(16\).
Moving the Constant First
When solving a quadratic equation by completing the square, it is often easiest to isolate the quadratic and linear terms first.
Consider
\[ x^2+4x-5=0. \]Add \(5\) to both sides:
\[ x^2+4x=5. \]Now complete the square.
Half of \(4\) is \(2\), and
\[ 2^2=4. \]Add \(4\) to both sides:
\[ x^2+4x+4=9. \]Then
\[ (x+2)^2=9. \]So
\[ x+2=\pm3, \]giving
\[ x=1 \]or
\[ x=-5. \]When the Leading Coefficient Is Not 1
Completing the square is simplest when the coefficient of \(x^2\) is \(1\).
Consider
\[ 2x^2+8x+3. \]Factor \(2\) from the quadratic and linear terms:
\[ 2(x^2+4x)+3. \]Complete the square inside the parentheses.
Half of \(4\) is \(2\), and \(2^2=4\):
\[ 2(x^2+4x+4-4)+3. \]Group:
\[ 2[(x+2)^2-4]+3. \]Distribute:
\[ 2(x+2)^2-8+3. \]Therefore,
\[ 2x^2+8x+3 = 2(x+2)^2-5. \]General Quadratic Form
A general quadratic expression has the form
\[ ax^2+bx+c, \qquad a\ne0. \]Factor \(a\) from the terms containing \(x\):
\[ a\left(x^2+\frac{b}{a}x\right)+c. \]Half of the coefficient of \(x\) inside the parentheses is
\[ \frac{b}{2a}. \]Its square is
\[ \frac{b^2}{4a^2}. \]So
\[ ax^2+bx+c = a\left(x+\frac{b}{2a}\right)^2 + c-\frac{b^2}{4a}. \]This is the general completed-square form of a quadratic expression.
Standard Form and Vertex Form
A quadratic function is often written in standard form:
\[ y=ax^2+bx+c. \]Completing the square converts it into vertex form:
\[ y=a(x-h)^2+k. \]The vertex is
\[ (h,k). \]Because a squared quantity is easy to interpret geometrically, vertex form often reveals the graph more clearly than expanded form.
Example: Finding the Vertex by Completing the Square
Consider
\[ y=x^2-6x+11. \]Complete the square:
\[ y=x^2-6x+9+2. \]Therefore,
\[ y=(x-3)^2+2. \]The vertex is
\[ (3,2). \]The expanded form made the coefficients easy to see. The completed-square form makes the graph's location easy to see.
Minimum and Maximum Values
Completed-square form can reveal an extremum immediately.
Suppose
\[ y=(x-4)^2+7. \]Since
\[ (x-4)^2\ge0, \]the smallest possible value of \(y\) is
\[ 7, \]which occurs when
\[ x=4. \]If instead
\[ y=-2(x+1)^2+5, \]then
\[ -2(x+1)^2\le0, \]so the maximum value is
\[ 5, \]occurring when
\[ x=-1. \]Completing the Square and Symmetry
For
\[ y=a(x-h)^2+k, \]the graph is symmetric about the vertical line
\[ x=h. \]Completing the square therefore exposes the axis of symmetry.
For the standard quadratic
\[ y=ax^2+bx+c, \]the completed-square formula shows that
\[ h=-\frac{b}{2a}. \]So the axis of symmetry is
\[ x=-\frac{b}{2a}. \]Geometric Interpretation
The phrase completing the square has a literal geometric interpretation.
Consider
\[ x^2+6x. \]Imagine an \(x\times x\) square with area
\[ x^2. \]The term \(6x\) can be split into two rectangles of area \(3x\):
\[ 6x=3x+3x. \]Placing one rectangle along each side leaves a missing corner that is a \(3\times3\) square.
Its area is
\[ 9. \]Adding that missing area creates a complete square:
\[ x^2+6x+9=(x+3)^2. \]The algebraic half-and-square rule comes directly from this geometry.
Completing the Square Versus Factoring
Factoring and completing the square are both equivalence transformations, but they reveal different structure.
For example,
\[ x^2-5x+6 \]factors as
\[ (x-2)(x-3), \]which makes the zeros visible.
Completing the square gives
\[ \left(x-\frac{5}{2}\right)^2-\frac{1}{4}, \]which makes the vertex and axis of symmetry visible.
The preferred form depends on the question being asked.
Completing the Square When Factoring Is Difficult
Some quadratics do not factor conveniently over the integers.
Consider
\[ x^2+4x+1=0. \]Move the constant:
\[ x^2+4x=-1. \]Add \(4\) to both sides:
\[ x^2+4x+4=3. \]Then
\[ (x+2)^2=3. \]So
\[ x+2=\pm\sqrt{3}, \]and
\[ x=-2\pm\sqrt{3}. \]Completing the square works even when simple integer factoring does not.
Deriving the Quadratic Formula
The quadratic formula can be derived by completing the square on the general equation
\[ ax^2+bx+c=0, \qquad a\ne0. \]Divide by \(a\):
\[ x^2+\frac{b}{a}x+\frac{c}{a}=0. \]Move the constant term:
\[ x^2+\frac{b}{a}x=-\frac{c}{a}. \]Add
\[ \left(\frac{b}{2a}\right)^2 = \frac{b^2}{4a^2} \]to both sides:
\[ x^2+\frac{b}{a}x+\frac{b^2}{4a^2} = -\frac{c}{a}+\frac{b^2}{4a^2}. \]The left side becomes
\[ \left(x+\frac{b}{2a}\right)^2. \]The right side becomes
\[ \frac{b^2-4ac}{4a^2}. \]So
\[ \left(x+\frac{b}{2a}\right)^2 = \frac{b^2-4ac}{4a^2}. \]Taking square roots gives
\[ x+\frac{b}{2a} = \pm \frac{\sqrt{b^2-4ac}}{2a}. \]Therefore,
\[ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]The quadratic formula is thus a general completed-square solution.
The Discriminant Appears Naturally
During the derivation, the expression
\[ b^2-4ac \]appears under the square root.
This quantity is called the discriminant.
Its sign determines the nature of the real solutions:
- If \(b^2-4ac>0\), there are two distinct real solutions.
- If \(b^2-4ac=0\), there is one repeated real solution.
- If \(b^2-4ac<0\), there are no real solutions, though complex solutions exist.
Completing the square therefore connects symbolic transformation directly to the solution structure of quadratics.
A Strategic View
Completing the square is useful when we want to:
- solve a quadratic equation;
- convert standard form to vertex form;
- identify a vertex or axis of symmetry;
- find a maximum or minimum value;
- derive the quadratic formula;
- rewrite expressions for integration or other advanced methods;
- recognize geometric or analytic structure hidden in an expanded expression.
It is one more example of a recurring algebraic principle: choose the equivalent form that exposes the structure you need.
Worked Example: Rewrite in Vertex Form
Rewrite
\[ y=x^2+10x+21 \]in vertex form.
Half of \(10\) is \(5\), and
\[ 5^2=25. \]Add and subtract \(25\):
\[ y=x^2+10x+25-25+21. \]Then
\[ y=(x+5)^2-4. \]The vertex is
\[ (-5,-4). \]Worked Example: Leading Coefficient Greater Than 1
Rewrite
\[ y=3x^2-12x+7 \]in completed-square form.
Factor \(3\) from the quadratic and linear terms:
\[ y=3(x^2-4x)+7. \]Half of \(-4\) is \(-2\), and
\[ (-2)^2=4. \]Add and subtract \(4\) inside the parentheses:
\[ y=3(x^2-4x+4-4)+7. \]Then
\[ y=3[(x-2)^2-4]+7. \]Distribute:
\[ y=3(x-2)^2-12+7. \]Therefore,
\[ y=3(x-2)^2-5. \]The vertex is
\[ (2,-5). \]Worked Example: Solve by Completing the Square
Solve
\[ x^2-8x+7=0. \]Move the constant:
\[ x^2-8x=-7. \]Half of \(-8\) is \(-4\), and
\[ (-4)^2=16. \]Add \(16\) to both sides:
\[ x^2-8x+16=9. \]Then
\[ (x-4)^2=9. \]Take square roots:
\[ x-4=\pm3. \]Therefore,
\[ x=7 \]or
\[ x=1. \]Worked Example: Irrational Solutions
Solve
\[ x^2+2x-5=0. \]Move the constant:
\[ x^2+2x=5. \]Half of \(2\) is \(1\), and
\[ 1^2=1. \]Add \(1\) to both sides:
\[ x^2+2x+1=6. \]Then
\[ (x+1)^2=6. \]So
\[ x+1=\pm\sqrt{6}, \]giving
\[ x=-1\pm\sqrt{6}. \]Common Errors
Forgetting to Square the Half
For
\[ x^2+10x, \]half of \(10\) is \(5\), but the number added is
\[ 5^2=25, \]not \(5\).
Adding to Only One Side of an Equation
When solving an equation, if \(9\) is added to the left side, it must also be added to the right side.
Ignoring the Leading Coefficient
For
\[ 2x^2+8x, \]first factor \(2\):
\[ 2(x^2+4x). \]Then complete the square inside the parentheses.
Forgetting the \(\pm\) While Solving
From
\[ (x-h)^2=k, \]with \(k>0\), solving for \(x\) requires
\[ x-h=\pm\sqrt{k}. \]Confusing Rewriting with Solving
Rewriting
\[ x^2+6x+5 \]as
\[ (x+3)^2-4 \]does not solve anything by itself. It simply expresses the same quadratic in a more revealing form.
Practice
1. Complete the Square
What number completes the square?
\[ x^2+14x+\square \]Check Answer
Half of \(14\) is \(7\), and
\[ 7^2=49. \]So
\[ x^2+14x+49=(x+7)^2. \]2. Negative Middle Term
Complete the square:
\[ x^2-18x+\square \]Check Answer
Half of \(-18\) is \(-9\), and
\[ (-9)^2=81. \]Therefore,
\[ x^2-18x+81=(x-9)^2. \]3. Rewrite a Quadratic
Rewrite in completed-square form:
\[ x^2+4x+9. \]Check Answer
\[ x^2+4x+9 = x^2+4x+4+5 = (x+2)^2+5. \]4. Solve by Completing the Square
Solve:
\[ x^2+6x-7=0. \]Check Answer
\[ x^2+6x=7. \]Add \(9\):
\[ (x+3)^2=16. \]Then
\[ x+3=\pm4. \]So
\[ x=1 \]or
\[ x=-7. \]5. Find the Vertex
Rewrite and identify the vertex:
\[ y=x^2-8x+13. \]Check Answer
\[ y=x^2-8x+16-3 \]so
\[ y=(x-4)^2-3. \]The vertex is
\[ (4,-3). \]6. Leading Coefficient
Rewrite in completed-square form:
\[ 2x^2+12x+5. \]Check Answer
\[ 2x^2+12x+5 = 2(x^2+6x)+5 \] \[ = 2(x^2+6x+9-9)+5 \] \[ = 2(x+3)^2-18+5 \] \[ = 2(x+3)^2-13. \]Key Ideas
- Completing the square rewrites a quadratic into an equivalent perfect-square form.
- For \(x^2+bx\), add \(\left(\frac{b}{2}\right)^2\) to create a perfect square trinomial.
- When rewriting an expression, add and subtract the same quantity to preserve its value.
- When solving an equation, add the same quantity to both sides to preserve equality.
- If the leading coefficient is not \(1\), factor it from the quadratic and linear terms first.
- Completed-square form reveals the vertex, axis of symmetry, and extrema of a quadratic function.
- Completing the square can solve quadratics that do not factor conveniently.
- The quadratic formula is derived by completing the square on a general quadratic equation.
- Taking square roots while solving an equation requires both positive and negative branches when appropriate.
- Completing the square is best understood as a structural equivalence transformation, not merely a memorized procedure.
Where This Leads
Completing the square shows how a carefully chosen equivalent form can expose both algebraic and geometric information.
Next, we will study Algebraic and Conditional Identities, distinguishing equations that are true throughout a domain from identities whose validity depends on explicit restrictions or conditions.