Solving quadratic equations by completing the square. Includes adding the square of half the coefficient of x to build a perfect square trinomial, solving an equation by completing the square and taking roots, deriving the quadratic formula by the same method, reading a vertex from the completed form, and choosing an appropriate solution method for a given equation.
Subject: Algebra 1 · 65 slides · symbolic lesson
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Title
Algebra 1 · Chapter 12 — Radicals and More Connections to Geometry
Completing the Square
Objectives
Five outcomes, each one you can test yourself on with a pencil and no answer key.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-722 — the lesson these objectives are drawn from
Warm-up
Lesson 10.7 recognised perfect square trinomials. This lesson builds one on purpose, by choosing the constant that makes the pattern fit.
Discussion prompt
What constant would make x squared plus eight x plus something a perfect square trinomial? What would it then factor as?
Hint: The middle term must be twice the product of the roots.
Answer:
\[ x^2 + 8x + 16 = (x + 4)^2 \]
Sixteen works, because twice x times four is eight x. Notice that four is half of eight and sixteen is four squared — that pattern is the whole technique, and this lesson turns it into a method for solving equations.
Concept
To complete the square of x squared plus b x, add the square of half the coefficient of x. The result is the square of x plus half that coefficient.
completing the square — Adding the square of half the coefficient of x to an expression of the form x squared plus b x, so that it becomes a perfect square trinomial.
Half of b appears inside the finished bracket.
Figure (svg): Halving the coefficient and squaring it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-716
Section
Section 1
Concept
For x squared plus b x, adding b over two, squared, produces a perfect square trinomial that factors as x plus b over two, all squared.
\[ x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2 \]
The halved coefficient appears in the bracket.
Figure (svg): Halving the coefficient and squaring it
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-716 — the Completing the Square rule and Example 1
Picture it
Two steps, one number.
Figure (svg): Halving the coefficient and squaring it
The same halved number does two jobs: squared it is the term to add, and unsquared it sits in the bracket. Computing it once saves work at both ends.
Worked example
This is Example 1 from the textbook.
\[ \text{What term completes the square of } x^2 + 8x? \]
Take the coefficient of x
Why: Read it off.
\[ 8 \]
Halve it
Why: Four.
\[ 4 \]
Square that
Why: Sixteen.
\[ 16 \]
Write the factored form
Why: The halved value goes inside.
\[ (x + 4) ^{2} \]
Figure (svg): Halving the coefficient and squaring it
\[ x^2 + 8x + 16 = (x + 4)^2 \]
Verify: expand the answer
Why: x plus four, squared, gives x squared plus eight x plus sixteen, which is the trinomial we built. Expanding the bracket is the natural check and takes one line.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-716
Faded example
Two steps.
Fill in the blanks
x^2 + 10x: \quad \text5 10 \text25 ___, \text___ ___
Why: The halved value five sits inside the finished bracket and its square twenty-five is the term added. Both come from the same halving.
Worked example
Guided Practice 1 to 4.
\[ \text{Complete the square of } x^2 + 2x, \; x^2 + 4x, \; x^2 + 6x \text{ and } x^2 + 10x. \]
Halve and square each
Why: One, two, three, five.
\[ 1, 4, 9, 25 \]
Note the pattern
Why: The added terms are squares.
Write the brackets
Why: Using the halved values.
\[ (x + 1) ^{2}, (x + 2) ^{2} \]
And the rest
Why: Three and five.
\[ (x + 3) ^{2}, (x + 5) ^{2} \]
Figure (svg): Halving the coefficient and squaring it
\[ 1, \; 4, \; 9, \; 25 \]
Verify: check one by expanding
Why: x plus five, squared, is x squared plus ten x plus twenty-five, matching the fourth. Every one of these can be checked the same way in a single line.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-716
Trap
\[ x^2 + 8x + 4 = (x + 4)^2 \]
Halve the coefficient and add that
Why: Four is the number that goes in the bracket, so four was added.
Four goes inside the bracket and its square goes into the trinomial. Expanding x plus four squared gives a constant of sixteen, not four, so the two sides do not match.
\[ x^2 + 8x + 16 = (x + 4)^2 \]
Halve, then square, and add the squared value
Why: Two operations, not one.
Expanding the bracket immediately shows which number belongs where.
Matching
Half the coefficient, squared.
Match the pairs
Why: Every added term is a perfect square, since it is the square of the halved coefficient. That is a quick check that the halving was done before the squaring.
Prediction
Such as x squared plus 5x.
Predict first
What term completes the square?
Correct: 25/4, the square of five halves.
\[ x^2 + 5x + \tfrac{25}{4} = \left(x + \tfrac{5}{2}\right)^2 \]
Why: The rule is the same whatever b is: halve it and square the result. Half of five is five halves, whose square is twenty-five quarters, and the trinomial factors as x plus five halves, all squared. Odd coefficients simply produce fractions, which is inconvenient rather than impossible — and it is one reason completing the square is often not the quickest method.
Socratic
The rule could be anything.
Discussion prompt
Explain why halving is the right operation, using the expansion of a squared binomial. Then connect it to the area model.
Hint: What is the middle term of x plus k, squared?
Answer:
Expanding x plus k, squared, gives x squared plus two k x plus k squared, so the middle coefficient is twice k. To match a middle coefficient of b, k must be half of b — and the constant that appears is then k squared, which is b over two, squared.
In the area model, the b x term is split into two rectangles of b over two each, one along each side of the x by x square. The corner they leave empty is a square of side b over two, so its area is that value squared. The picture and the algebra give the same number for the same reason.
Section
Section 2
Concept
Adding the completing term to both sides of an equation turns one side into a perfect square. The equation is then in the form Lesson 9.2 solved by taking square roots.
The term must be added to both sides, not just one.
Figure (svg): A quadratic equation solved by completing the square
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 717-717 — Example 2 and its Study Tip on adding to both sides
Picture it
Complete, square, root, solve.
Figure (svg): A quadratic equation solved by completing the square
The plus-or-minus enters at the rooting step, exactly as in Lesson 9.2. Everything before it exists to get the equation into that shape.
Worked example
This is Example 2 from the textbook.
\[ \text{Solve } x^2 + 10x = 24 \text{ by completing the square.} \]
Find the completing term
Why: Half of ten, squared.
\[ 25 \]
Add it to both sides
Why: Keeping the equation balanced.
\[ x ^{2} + 10 x + 25 = 49 \]
Write as a square
Why: The halved value inside.
\[ (x + 5) ^{2} = 49 \]
Take roots and solve
Why: Both signs, then subtract five.
\[ x = 2 \text{ or } -12 \]
Figure (svg): A quadratic equation solved by completing the square
\[ x = 2 \text{ and } x = -12 \]
Verify: check both in the original
Why: At two, four plus twenty is twenty-four. At negative twelve, a hundred and forty-four minus a hundred and twenty is also twenty-four. Both satisfy the equation as first written.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 717-717
Faded example
The equation stays balanced.
Fill in the blanks
x^2 + 10x = 24 \;\to\; x^2 + 10x + 25 = 24 + 25 = 49
Why: Whatever is added to one side must be added to the other, or the equation describes something different. Forty-nine happens to be a perfect square here, which makes the rooting tidy.
Worked example
Guided Practice 5, where a constant must be moved.
\[ \text{Solve } x^2 + 2x - 3 = 0 \text{ by completing the square.} \]
Move the constant
Why: Add three to both sides.
\[ x ^{2} + 2 x = 3 \]
Complete the square
Why: Half of two, squared.
\[ x ^{2} + 2 x + 1 = 4 \]
Write as a square
Why: One inside the bracket.
\[ (x + 1) ^{2} = 4 \]
Take roots
Why: Both signs, then subtract one.
\[ x = 1 \text{ or } -3 \]
Figure (svg): A quadratic equation solved by completing the square
\[ x = 1 \text{ and } x = -3 \]
Verify: compare with factoring
Why: The original factors as x plus three times x minus one, giving the same solutions in one line. Completing the square worked but was the longer route here, which is a point the last section returns to.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 717-717
Error analysis
The student completed the square on an equation.
Annotate
On: \( \begin{aligned} x^2 + 10x &= 24 \\ x^2 + 10x + 25 &= 24 \\ (x + 5)^2 &= 24 \end{aligned} \)
Completing the square is the one method in this course where a term is deliberately added to an expression, so the balancing has to be conscious. Every other technique only ever rearranges what is already there.
Sorting
The order of the method.
Sort into buckets
Sort each action by whether it comes before or after the square is written.
The factoring step is the hinge: everything before it prepares the square and everything after it is Lesson 9.2's rooting method. Recognising that split makes the procedure easier to remember.
Elimination
For x squared + 10x = 24.
Eliminate the wrong options
Which term completes the square?
Survives elimination: A
Why: Half of ten is five and five squared is twenty-five. Each wrong option skips one of the two steps, and expanding x plus five squared confirms which constant belongs.
Socratic
Factoring sometimes fails.
Discussion prompt
Explain why completing the square can solve any quadratic equation, even one that does not factor. Then say what the answer looks like in that case.
Hint: What does the method require of the coefficients?
Answer:
The method needs only that half the coefficient of x can be squared, which is always possible whatever the numbers are. It does not depend on finding integers with a particular sum and product, which is exactly what factoring can fail to do.
When the equation does not factor over the integers, the number on the right after completing is not a perfect square, so taking roots leaves a radical — and the answers come out as expressions like negative five plus or minus the root of thirty. That is a perfectly good exact answer, and it is precisely the form the quadratic formula produces.
Section
Section 3
Concept
Completing the square on the general equation a x squared plus b x plus c equals nought produces the quadratic formula. Every step is one already used on numerical equations.
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
The formula is a result rather than an assumption.
Figure (svg): The quadratic formula derived by completing the square
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 717-717 — Example 3, Develop the Quadratic Formula
Picture it
Divide, complete, root, solve.
Figure (svg): The quadratic formula derived by completing the square
The formula memorised in Lesson 9.6 is the output of this derivation. Seeing where it comes from makes it something you could rebuild rather than only recall.
Worked example
This is Example 3 from the textbook.
\[ \text{Complete the square on } ax^2 + bx + c = 0. \]
Move c and divide by a
Why: The leading coefficient must be one.
\[ x^2 + \tfrac{b}{a}x = -\tfrac{c}{a} \]
Complete the square
Why: Half of b over a, squared.
\[ +\tfrac{b^2}{4a^2} \text{ both sides} \]
Write the left as a square
Why: The halved value inside.
\[ \left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2} \]
Take roots and solve
Why: Both signs, then subtract.
\[ x = \tfrac{-b \pm \sqrt{b^2-4ac}}{2a} \]
Figure (svg): The quadratic formula derived by completing the square
\[ x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Verify: check the right side's combination
Why: The right side became b squared over four a squared minus c over a, and putting those over the common denominator four a squared gives b squared minus four a c over four a squared. Taking its root gives the radical over two a, which is where the formula's denominator comes from.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 717-717
Faded example
The leading coefficient must be one.
Fill in the blanks
ax^2 + bx = -c \;\to\; x^2 + \tfraca2a}x = -\tfrac______ \;\to\; \text___ \left(\tfrac______}\right)^2
Why: After dividing, the coefficient of x is b over a, and half of that is b over two a. Squaring it gives the term to add, and the unsquared version appears in the final bracket.
Worked example
Reading the derivation for meaning.
\[ \text{At which step does } b^2 - 4ac \text{ first appear, and why does its sign matter?} \]
Find the step
Why: Combining the right side.
\[ \tfrac{b^2 - 4ac}{4a^2} \]
Note what happens next
Why: A square root is taken.
Consider a negative value
Why: The numerator would be negative.
Connect to Lesson 9.7
Why: The count of solutions.
Figure (svg): The quadratic formula derived by completing the square
\[ \left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2} \]
Verify: see why the sign decides the count
Why: The left side is a square, which cannot be negative, so if the right side is negative there is no real solution. If it is nought the square is nought and there is one, and if positive there are two. That is exactly Lesson 9.7's three cases, now with a reason behind them.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 717-717
Trap
\[ ax^2 + bx = -c \;\Longrightarrow\; \text{add } \left(\tfrac{b}{2}\right)^2 \]
Complete the square with the coefficient as it stands
Why: The rule says half the coefficient of x, and b is that coefficient.
The completing rule applies to x squared plus b x, with a leading coefficient of one. With a in front, the coefficient to halve is b over a, so the leading coefficient must be divided out first.
\[ x^2 + \tfrac{b}{a}x = -\tfrac{c}{a} \;\Longrightarrow\; \text{add } \left(\tfrac{b}{2a}\right)^2 \]
Divide through by a before completing anything
Why: The rule assumes a leading coefficient of one.
This is the step that produces the two a in the formula's denominator.
Matching
Where each piece comes from.
Match the pairs
Why: Every feature of the formula traces back to one step of the derivation. That is why rebuilding it is possible even if the formula itself is half-remembered.
Hypothesis
The formula can simply be memorised.
Predict first
What does the derivation provide that memorising does not?
Correct: It explains where every part of the formula comes from, including the discriminant.
Lesson 9.7's three cases follow immediately from the step before the square root.
Why: The two a in the denominator comes from dividing by a, the negative b from the halved coefficient, and the discriminant from combining the right side just before the root is taken. That last point also explains why the discriminant's sign decides the number of solutions: a square cannot be negative. Memorising gives none of that, and a half-remembered formula cannot be repaired without it.
Socratic
The derivation makes it visible.
Discussion prompt
Use the step before the square root to explain the three cases of Lesson 9.7. Then say why this is more satisfying than the rule as stated there.
Hint: What kind of quantity is the left side?
Answer:
At that step the equation reads a squared bracket equal to the discriminant over four a squared. A square is never negative, so if the discriminant is negative no real x can satisfy the equation; if it is nought the bracket must be nought, giving one value; and if positive the bracket can be either root, giving two.
Lesson 9.7 stated those three outcomes as a rule to be applied, which works but explains nothing. Here they follow from a single fact about squares, so the rule becomes a consequence rather than something extra to remember — and that is generally what a derivation buys.
Section
Section 4
Concept
Writing a quadratic function as a square plus a constant shows its vertex directly, because a square is smallest when the expression inside it is nought.
\[ y = (x + 5)^2 - 49 \;\Longrightarrow\; \text{vertex } (-5, -49) \]
A second payoff of the same technique.
Figure (svg): A quadratic rewritten so its vertex can be read off
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-722 — the completed square form and its connection to the graph
Picture it
Read it off the brackets.
Figure (svg): A quadratic rewritten so its vertex can be read off
Lesson 9.4 found vertices with negative b over two a. This form gives the same point with no formula at all, which is why it is worth being able to produce.
Worked example
Applying the technique to a function rather than an equation.
\[ \text{Write } y = x^2 + 10x - 24 \text{ in completed form and find its vertex.} \]
Complete the square
Why: Half of ten, squared.
\[ \text{add and subtract } 25 \]
Group the square
Why: The first three terms.
\[ (x ^{2} + 10 x + 25) - 25 - 24 \]
Write it as a square
Why: Five inside.
\[ (x + 5) ^{2} - 49 \]
Read the vertex
Why: The bracket is nought at x equal to negative five.
\[ (-5, -49) \]
Figure (svg): A quadratic rewritten so its vertex can be read off
\[ y = (x + 5)^2 - 49, \quad \text{vertex } (-5, -49) \]
Verify: check against the formula from Lesson 9.4
Why: Negative b over two a is negative ten over two, which is negative five, and substituting gives twenty-five minus fifty minus twenty-four, or negative forty-nine. Both methods agree, as they must.
Faded example
Where the bracket is nought.
Fill in the blanks
y = (x + 5)^2 - 49: \quad \text-5 x = -49, \; y = ___
Why: The x-coordinate flips sign from the bracket and the y-coordinate is the constant as written. Those two conventions differ, which is the usual source of confusion here.
Worked example
The reasoning behind reading it off.
\[ \text{Why is the vertex of } y = (x + 5)^2 - 49 \text{ at } x = -5? \]
Look at the square
Why: It is never negative.
\[ (x + 5)^2 \ge 0 \]
Find its smallest value
Why: When the bracket is nought.
\[ x = -5 \]
Find y there
Why: Nought minus forty-nine.
\[ -49 \]
Conclude
Why: That is the lowest point.
Figure (svg): A quadratic rewritten so its vertex can be read off
\[ (x+5)^2 = 0 \;\Longrightarrow\; x = -5 \]
Verify: test either side
Why: At negative four the square is one and y is negative forty-eight; at negative six it is also one and y is negative forty-eight. Both are above negative forty-nine, confirming it as the minimum.
Trap
\[ y = (x + 5)^2 - 49 \;\Longrightarrow\; \text{vertex } (5, -49) \]
Read the five from the bracket
Why: The number five is right there.
The vertex is where the bracket is nought, and x plus five is nought when x is negative five. The sign flips, exactly as it did when reading intercepts from factors in Lesson 10.4.
\[ \text{vertex } (-5, -49) \]
Solve for the value making the bracket nought
Why: Rather than copying the constant.
The y-coordinate, by contrast, is read directly with its own sign.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Standard form | Completed form | |
|---|---|---|
| How to get x | compute -b over 2a | read it from the bracket, with the sign flipped |
| How to get y | substitute that x back in | read the constant outside |
| Work required | a formula and a substitution | completing the square first |
Neither is free: one needs a formula and a substitution, the other needs the completing done first. The completed form pays off when several features are wanted at once.
Elimination
For y equal to (x - 3) squared + 2.
Eliminate the wrong options
Where is the turning point?
Survives elimination: A
Why: The bracket vanishes at x equal to three and the constant outside is two. The x-coordinate flips the bracket's sign and the y-coordinate does not flip anything, which is the distinction to hold on to.
Socratic
Standard form does not.
Discussion prompt
Explain why a square plus a constant makes the turning point visible. Then say what the same form shows about the range.
Hint: What is the smallest a square can be?
Answer:
A square is never negative and is nought exactly when its bracket is, so the whole expression is smallest at that one value of x. Everywhere else the square contributes something positive, pushing the value up — which is precisely what a minimum turning point means.
The same reasoning gives the range immediately: the output is the constant outside plus something non-negative, so it is at least that constant. For y equal to x plus five squared minus forty-nine the range is y at least negative forty-nine, which standard form would not reveal without further work.
Section
Section 5
Concept
A quadratic equation can be solved by taking roots, factoring, the quadratic formula or completing the square. The best choice depends on the shape of the equation.
Completing the square is rarely the quickest.
Figure (svg): Two columns comparing when to complete the square with when to use another method
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 718-718 — Example 4, Choose a Solution Method, on the penguin's leap
Picture it
Shape decides.
Figure (svg): Two columns comparing when to complete the square with when to use another method
Completing the square always works and is seldom fastest, which is a fair summary of its role. Its importance lies in what it derives rather than in what it solves.
Worked example
This is Example 4 from the textbook.
\[ \text{With } h = -0.05x^2 + 1.178x, \text{ find where the penguin re-enters the water.} \]
Set the height to nought
Why: Water level.
\[ -0.05 x ^{2} + 1.178 x = 0 \]
Notice there is no constant
Why: Every term has an x.
Factor
Why: The zero-product property applies.
\[ x(-0.05 x + 1.178) = 0 \]
Solve
Why: Nought, or the other factor.
\[ x \approx 23.6 \]
Figure (svg): The arc of a leaping penguin
\[ x = 0 \text{ or } x \approx 23.6 \text{ feet} \]
Verify: check the arithmetic
Why: Dividing 1.178 by 0.05 gives 23.56, and substituting confirms the height is nought there. The solution at nought is where the penguin left the water, so the leap covers the distance between the two.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 718-718
Sorting
Look at the shape of the equation.
Sort into buckets
Sort each equation by the method that suits it.
Only one of the six genuinely needs the general method. Checking the discriminant first is the quickest way to know which situation you are in.
Worked example
Comparing the available methods.
\[ \text{Why not complete the square on } -0.05x^2 + 1.178x = 0? \]
Check the leading coefficient
Why: Not one, and not tidy.
\[ -0.05 \]
Imagine dividing through
Why: The coefficient of x becomes awkward.
\[ -23.56 \]
Halve and square that
Why: A messy number.
\[ \approx 138.8 \]
Compare with factoring
Why: One line, exact.
Figure (svg): Two columns comparing when to complete the square with when to use another method
\[ x(-0.05x + 1.178) = 0 \]
Verify: confirm both give the same answer
Why: Completing the square would eventually produce the same 23.56, since all four methods are exact. The difference is entirely in the effort, and a missing constant term is the clearest possible signal to factor.
Trap
Always use the quadratic formula, since it always works.
Apply the general method regardless of the equation
Why: It never fails, so it seemed the safe choice.
For an equation with no constant term, factoring takes one line and the formula takes five. Reaching for the general method every time is reliable and often several times more work than necessary.
Look at the equation's shape first, then choose.
Spend two seconds deciding before starting
Why: No constant term, or easy factors, or neither.
The formula remains the fallback when nothing else fits.
Faded example
Every term contains x.
Fill in the blanks
-0.05x^2 + 1.178x = 0 \;\to\; x(-0.05x + 1.178) = 0 \;\to\; x = 0 \text1.178 x = 23.6 \div 0.05 \approx ___
Why: A missing constant term means x is a factor of the whole expression, so nought is always one solution. That signal makes factoring the obvious first move.
Prediction
For a quadratic equation equal to nought.
Predict first
What can you say immediately?
Correct: Zero is one of the solutions.
\[ ax^2 + bx = 0 \;\Longrightarrow\; x(ax + b) = 0 \]
Why: With no constant term, every term contains x, so x can be factored out and the zero-product property gives x equal to nought as one solution. It also means the equation always factors, since one factor is x itself, so the general method is never needed. Recognising this takes a glance and saves the whole calculation.
Socratic
It is rarely the fastest method.
Discussion prompt
Give two reasons for learning a technique that is seldom the quickest way to solve an equation. Then say where it becomes essential.
Hint: Think about what it produces rather than what it solves.
Answer:
First, it derives the quadratic formula, so the formula becomes a result you could rebuild rather than a string of symbols to recall. Second, it produces the completed form, which shows a parabola's vertex and range directly and is the standard way of writing such a function in later courses.
It becomes essential wherever the completed form itself is wanted rather than the solutions — sketching a graph from its turning point, finding a maximum or minimum value, or working with the equations of circles and other curves, all of which are built on exactly this rearrangement. Its role is structural rather than computational, which is why it is worth the effort despite rarely winning a race.
Comparison
Fill the blanks from memory before you scroll back.
Comparison matrix
| Method | Best when | Limitation |
|---|---|---|
| Taking square roots | there is no x term | only works for that shape |
| Factoring | the discriminant is a perfect square | fails when it is not |
| Completing the square | the leading coefficient is 1 | awkward fractions otherwise |
The quadratic formula is the missing fourth row: it always works and is never especially quick. Every other method is faster in the case it was designed for.
Pattern
To solve a quadratic equation by completing the square, these five moves cover it.
Step three is the only new one, and step four is where the method hands over to Lesson 9.2's rooting technique.
OpenStax Elementary Algebra 2e, §10.2 Solve Quadratic Equations by Completing the Square §10.2
Check
Halve, then square.
Check your understanding
What term completes the square of x squared + 8x?
Answer: A
Why: Half of eight is four and four squared is sixteen, giving x plus four, all squared.
Check
Add to both sides.
Check your understanding
Solve x squared + 10x = 24 by completing the square.
Answer: A
Why: Adding twenty-five to both sides gives x plus five squared equal to forty-nine, so x plus five is plus or minus seven.
Check
Read the bracket.
Check your understanding
What is the vertex of y = (x + 5) squared - 49?
Answer: A
Why: The bracket is nought when x is negative five, and the value there is negative forty-nine.
Real world
This is the penguin question from the lesson opener. A leaping penguin's path is modelled by h equal to negative 0.05 x squared plus 1.178 x, with h the height above the water and x the horizontal distance.
Discussion prompt
Find how far the penguin travels before re-entering the water, choose your method deliberately, and say why completing the square would have been the wrong choice here.
Hint: Set the height to nought and look at the terms.
Answer:
\[ -0.05x^2 + 1.178x = 0 \;\Longrightarrow\; x(-0.05x + 1.178) = 0 \;\Longrightarrow\; x \approx 23.6 \text{ feet} \]
There is no constant term, so x factors out of every term and the zero-product property finishes the problem in one line. The solution at nought is where the penguin left the water, and 23.6 feet is where it comes back down.
Completing the square would have required dividing by negative 0.05 first, turning the coefficient of x into about negative 23.56, then halving and squaring that to get a constant near a hundred and thirty-nine — several lines of awkward decimals to reach the same answer. All four methods are exact, so the choice is entirely about effort, and a missing constant term is the clearest signal there is to factor.
Commit first
Answer, then rate your confidence honestly. Confident and wrong is the combination worth finding.
Predict first
To solve x squared + 10x = 24 by completing the square, what should be added?
Correct: 25, to both sides.
\[ x^2 + 10x + 25 = 24 + 25 = 49 \]
Why: Half of ten is five and five squared is twenty-five, so twenty-five is the completing term — and it must go on both sides or the equation is no longer the one you started with. Adding it to the left alone changes the solutions entirely and is the error the textbook's Study Tip warns against explicitly, since completing the square is the one method here that deliberately introduces a new term rather than merely rearranging. Adding five instead skips the squaring: five is the number that ends up inside the bracket, not the one added. Adding a hundred squares the coefficient without halving it first. With twenty-five on both sides the equation becomes x plus five squared equal to forty-nine, giving x equal to two or negative twelve, and both check.
Explain it
They added the completing term to one side only.
Discussion prompt
In no more than four sentences, explain why both sides need it. Then give them a way to check they have not unbalanced the equation.
Hint: What does an equation assert?
Answer:
A usable answer: an equation says two things are equal, so anything done to one side must be done to the other or the statement stops being true. Adding twenty-five to the left alone gives a different equation with different solutions.
The check is to substitute your answers back into the original equation. If you unbalanced it, the answers will satisfy your rearranged version but not the one you were actually asked about.
Exit ticket
Name the weakest spot before you close the deck. That is the one worth ten minutes tonight.
Predict first
Which of these would you least want to be handed cold on a quiz tomorrow?
Correct: Whichever you picked is the right answer — and each one has a specific fix.
Why: The completing term is fixed by halving before squaring, in that order. Balancing is fixed by writing both additions on the same line. The derivation is fixed by working through it once with the numerical version beside it for comparison. Method choice is fixed by looking for a missing x term, a missing constant, or an easy factorisation before starting. Pick yours and do five of that kind tonight rather than twenty mixed ones.
Connect it up
Do this on paper. It is worth more than rereading the slides.
Draw it
At the top of a page draw the area model for completing the square of x squared plus eight x, labelling the two rectangles of four x and the empty corner, and write beside it the halve-then-square rule. Underneath, complete the square for four expressions, including one with an odd coefficient so the fraction appears, and expand each finished bracket to check. In the middle, solve an equation by completing the square, writing the addition to both sides on a single line so the balance is visible, and check both solutions in the original. Beneath that, carry out the derivation of the quadratic formula with the general letters, and beside each line write which part of the final formula that step produced. In the lower half, write a quadratic in completed form, read its vertex and range off directly, and confirm the vertex with negative b over two a. Finally, in the margin, list the four solution methods with the shape of equation each one suits.
Your derivation's lines should correspond one to one with the numerical example above it. If a step in the general version has no counterpart in the numerical one, something has been skipped in whichever is shorter.
Recap
Five things, and the third is why the technique matters.
| If the question says | Your first move is |
|---|---|
| Complete the square | Halve the coefficient, then square it |
| Solve by completing the square | Divide by a, then move the constant |
| Add the completing term | Put it on both sides |
| Find the vertex | Complete the square and read the bracket |
| Solve any quadratic | Look at its shape before choosing a method |
Lesson 12.6 turns to geometry. The Pythagorean theorem relates the sides of a right triangle through a sum of squares, and the square roots of Chapter 9 are what recover a side length from it.
McDougal Littell Algebra 1: Concepts and Skills, Ch. 12 Radicals and More Connections to Geometry — Lesson 12.5 Completing the Square §12.5, pp. 716-722 — everything on these slides traces back here
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