Solving when one side is already a perfect square, the number that completes x squared plus bx, solving any quadratic equation by completing the square with any leading coefficient, an area model, and rewriting a function in vertex form to read off its maximum.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 4 — Quadratic Functions and Factoring
Complete the Square
Objectives
Five outcomes. The third makes every quadratic solvable; the fifth is the bonus that comes free.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 284-289 — the lesson these objectives are drawn from
Warm-up
Lesson 4.5 could solve a square equal to a number. Lessons 4.3 and 4.4 could factor. Neither handles what follows.
Discussion prompt
Try to solve x squared minus 12x plus 4 equals zero, first by factoring, then by taking square roots. What goes wrong in each case?
Hint: For factoring, list the pairs whose product is 4. For roots, look for a first-power term.
Answer:
\[ x^2 - 12x + 4 = 0 \]
Factoring fails: the pairs multiplying to 4 are 1 and 4, 2 and 2, and their negatives, and none sums to negative 12. Taking roots fails too, because there is a first-power term, so the left side is not a single square.
The equation does have solutions — the parabola crosses the axis twice. What is missing is a method, and this lesson supplies one that never fails.
Concept
Lesson 4.5's method needs one side to be a square. Completing the square adds a carefully chosen number to both sides to force the left side into that shape, after which the old method finishes the job. Since the number can always be found, the method works on every quadratic equation.
completing the square — The process of adding a constant to an expression x squared plus bx so that it becomes a perfect square trinomial. The constant is half of b, squared.
\[ x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2 \]
The name is literal. Drawn as areas, x squared plus bx is a square with a rectangle attached; splitting the rectangle in two and moving half of it produces a shape that is a square except for one missing corner, and the number added is exactly that corner.
Figure (svg): Two area diagrams showing an incomplete square of side x plus half b, and the corner square that completes it
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 284-284
Section
Section 1
Concept
If one side of an equation is a perfect square trinomial, write it as a binomial squared and the equation becomes a square equal to a number — exactly the shape Lesson 4.5 solves. No new technique is needed.
\[ x^2 - 8x + 16 = 25 \;\Longrightarrow\; (x-4)^2 = 25 \;\Longrightarrow\; x = 4 \pm 5 \]
The perfect-square test from Lesson 4.3 is what spots it: both outer terms are squares and the middle term is exactly twice the product of their roots.
Figure (svg): An equation whose left side is already a perfect square, solved in four lines
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 284-284 — Solve a quadratic equation by finding square roots
Picture it
Example 1, from the trinomial to the two solutions.
Figure (svg): An equation whose left side is already a perfect square, solved in four lines
The rest of the lesson exists to produce that second line for equations that do not arrive with it already written.
Worked example
Example 1. Every step is from Lessons 4.3 and 4.5.
\[ \text{Solve } x^2 - 8x + 16 = 25. \]
Test the left side for the perfect-square shape
Why: Sixteen is 4 squared and 8x is twice x times 4, with a minus sign, so the pattern applies.
\[ (x - 4) ^{2} = 25 \]
Take square roots of both sides
Why: Twenty-five is positive, so there are two real roots and both signs are written.
\[ x - 4 = +- 5 \]
Isolate x
Why: Adding 4 to both sides gives 4 plus or minus 5.
\[ x = 4 + - 5 \]
Write out the two solutions
Why: Four plus 5 is 9; 4 minus 5 is negative 1.
\[ x = 9\text{ or } x = -1 \]
Figure (svg): The solution to Worked example a square handed to you shown as a ladder of expressions, one row per algebraic move
\[ x = 9 \quad \text{or} \quad x = -1 \]
Verify: solve it a second way
Why: Rewriting as x squared minus 8x minus 9 equals zero and factoring gives the quantity x minus 9, times the quantity x plus 1, whose roots are 9 and negative 1. The two routes agree, which is what the textbook's Another Way note points out — and it is a reminder that having two methods lets each check the other.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 284-284
Matching
Halve the middle coefficient to find what goes inside.
Match the pairs
Why: In every case the number inside the bracket is half the middle coefficient, sign included, and the constant is that number squared. Both relationships are worth noticing now, because the next idea runs them in reverse: given the middle coefficient, produce the constant.
Worked example
Guided Practice 1 to 3.
\[ \text{Solve } x^2 + 6x + 9 = 36, \; x^2 - 10x + 25 = 1, \; x^2 - 24x + 144 = 100. \]
First: recognise and take roots
Why: Nine is 3 squared and 6x is twice x times 3, so the bracket is x plus 3, and the root of 36 is 6.
\[ x = -3 + - 6 \]
Second
Why: Twenty-five is 5 squared and 10x is twice x times 5, and the root of 1 is 1.
\[ x = 5 + - 1 \]
Third
Why: One hundred forty-four is 12 squared and 24x is twice x times 12, and the root of 100 is 10.
\[ x = 12 + - 10 \]
Write out all six solutions
Why: Three, negative 9; then 6, 4; then 22, 2.
\[ 3\text{ and } -9; 6\text{ and } 4; 22\text{ and } 2 \]
Figure (svg): The solution to Worked example three more perfect squares shown as a ladder of expressions, one row per algebraic move
\[ \{3, -9\}; \quad \{6, 4\}; \quad \{22, 2\} \]
Verify: check one pair against the vertex
Why: For the second, the two roots 6 and 4 average to 5, which is the number inside the bracket — as it must be, since the roots are symmetric about the value that makes the bracket vanish. That symmetry is a fast check on every answer of this shape.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 285-285
Trap
\[ x^2 - 8x + 16 = 25 \]
Factor the left side and use the zero product property
Why: The left side does factor, as the quantity x minus 4, squared.
\[ (x-4)^2 = 25 \;\Longrightarrow\; x - 4 = 0 \;\Longrightarrow\; x = 4 \quad \text{(wrong)} \]
Substituting 4 gives 16 minus 32 plus 16, which is 0, not 25. The zero product property was applied to a product equal to 25.
\[ (x-4)^2 = 25 \;\Longrightarrow\; x - 4 = \pm 5 \]
Take square roots, because the right side is not zero
Why: The zero product property needs a zero; taking roots needs only a number, and there is one here.
\[ x = 4 \pm 5 \;\Longrightarrow\; x = 9 \text{ or } x = -1 \]
Both methods are available once the equation is in standard form, but only one is available before that. Reading the right side first decides which one you are entitled to use.
Fill the middle
Example 1, at the last step.
Fill in the blanks
(x - 4)^2 = 25 \;\Longrightarrow\; x - 4 = \pm 5 \;\Longrightarrow\; x = 4 \pm 5
Why: Adding 4 to both sides gives 4 plus or minus 5, so the solutions are 9 and negative 1. Note that the 4 comes out with a plus sign even though the bracket read x minus 4 — the sign flips when it moves across the equals sign.
Sorting
Test the middle term against twice the product of the roots.
Sort into buckets
Sort each trinomial.
The two failures are the ones this lesson is built for: neither is a square, but a different constant would make each one into one.
Prediction
Commit before reasoning.
Predict first
The equation reads a bracket squared equals a positive number. How many real solutions?
Correct: Two, symmetric about the value that makes the bracket zero.
\[ (x-4)^2 = 25 \;\Longrightarrow\; x = 4 \pm 5 \quad \text{(two, centred on 4)} \]
Why: Taking roots gives the bracket equal to plus or minus the root, so the variable ends up the same distance either side of the number that makes the bracket vanish. The size of the number does not matter and neither does whether it is a perfect square — a non-square just makes the answers irrational rather than fewer. Only a negative right side reduces the count, and Lesson 4.6 handles that case with i.
Section
Section 2
Concept
To turn x squared plus bx into a perfect square trinomial, add the square of half of b. The resulting trinomial factors as the quantity x plus half of b, squared, and the number inside the bracket is the halving, not the squaring.
\[ x^2 + bx + \left(\tfrac{b}{2}\right)^2 = \left(x + \tfrac{b}{2}\right)^2 \]
The rule works whatever b is — negative, odd, fractional. For an odd b the constant comes out as a fraction, which is normal and not a sign that something has gone wrong.
Figure (svg): The three steps for finding the number that completes the square, applied to a middle coefficient of sixteen
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 284-285 — Completing the Square
Picture it
The area diagram the rule comes from.
Figure (svg): Two area diagrams showing an incomplete square of side x plus half b, and the corner square that completes it
The rectangle of area bx is split into two strips of width half of b, one moved beneath the square. The empty corner is a square of side half of b, so its area is half of b, squared — the number to add.
Worked example
Example 2, in the book's three steps.
\[ \text{Find } c \text{ so that } x^2 + 16x + c \text{ is a perfect square trinomial.} \]
Step 1: halve the coefficient of x
Why: Sixteen divided by 2 is 8.
\[ \frac{16}{2} = 8 \]
Step 2: square that result
Why: Eight squared is 64.
\[ 8 ^{2} = 64 \]
Step 3: that is c
Why: Replacing c with 64 gives the trinomial.
\[ x ^{2} + 16 x + 64 \]
Write it as a binomial squared
Why: The number inside the bracket is the Step 1 answer, 8, not the Step 2 answer.
\[ (x + 8) ^{2} \]
Figure (svg): The solution to Worked example make a perfect square trinomial shown as a ladder of expressions, one row per algebraic move
\[ c = 64, \qquad x^2 + 16x + 64 = (x+8)^2 \]
Verify: expand the binomial
Why: The quantity x plus 8, squared, is x squared plus 8x plus 8x plus 64, which is x squared plus 16x plus 64. The middle term appears twice, which is exactly why halving is the right first move: the 8 has to serve as both of the two 8x terms.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 285-285
Fill the middle
Guided Practice 4.
Fill in the blanks
x^2 + 14x + 49 = (x + 7)^2
Why: Half of 14 is 7, and 7 squared is 49. Notice that the 7 appears inside the bracket and the 49 outside it, which is the relationship worth carrying into the next idea: the number you add is the square of the number that ends up in the bracket.
Worked example
Guided Practice 4 to 6. The third is the interesting one.
\[ \text{Complete } x^2 + 14x + c, \; x^2 + 22x + c, \; x^2 - 9x + c. \]
First: halve and square
Why: Fourteen halves to 7, and 7 squared is 49.
\[ c = 49; (x + 7) ^{2} \]
Second: the same
Why: Twenty-two halves to 11, and 11 squared is 121.
\[ c = 121; (x + 11) ^{2} \]
Third: an odd coefficient
Why: Negative 9 halves to negative nine halves, which is not an integer.
\[ \text{half is } -\frac{9}{2} \]
Square the fraction anyway
Why: Negative nine halves squared is 81 over 4, so c is 81 over 4.
\[ c = \frac{81}{4}; (x - \frac{9}{2}) ^{2} \]
Figure (svg): The solution to Worked example three more, including an odd one shown as a ladder of expressions, one row per algebraic move
\[ (x+7)^2, \quad (x+11)^2, \quad \left(x - \tfrac{9}{2}\right)^2 \]
Verify: expand the fractional one
Why: The quantity x minus nine halves, squared, is x squared minus 9x plus 81 over 4. The middle term is right and so is the constant. A fractional constant is completely normal — it is the price of an odd middle coefficient, and it is why completing the square sometimes looks messier than factoring even though it always works.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 285-285
Error analysis
A student completes the square on the expression x squared plus 16x.
Annotate
On: \( x^2 + 16x + c: \quad 16^2 = 256, \text{ then } 256/2 = 128, \text{ so } c = 128 \)
Halve first, always. Expanding the resulting bracket takes five seconds and catches the reversal every time.
Matching
Halve, then square.
Match the pairs
Why: The three even coefficients give whole-number constants and the odd one gives a fraction. The sign of the middle coefficient never affects the constant, because squaring removes it — negative 9x and positive 9x both call for 81 over 4.
Ranking
Completing the square on an expression.
Put in order
Why: The last step is where most sign errors happen: the number inside the bracket is the output of step two, complete with its sign, not the output of step three. Writing the bracket immediately after adding the constant keeps the two straight.
Prediction
Commit before reasoning.
Predict first
For x squared minus 9x, what constant completes the square?
Correct: Eighty-one over 4, from half of negative 9, squared.
\[ x^2 - 9x + \tfrac{81}{4} = \left(x - \tfrac{9}{2}\right)^2 \]
Why: Half of negative 9 is negative nine halves, and its square is 81 over 4. Nothing requires the constant to be a whole number, and 81 over 4 is exact rather than an approximation — it is 20.25 written properly. This is precisely why completing the square works on every quadratic while factoring does not: fractions are allowed at every step.
Section
Section 3
Concept
To solve a quadratic equation by completing the square, first divide every term by the coefficient of the squared term so it becomes 1, then move the constant to the other side, add half the middle coefficient squared to both sides, and take roots.
\[ x^2 - 12x + 4 = 0 \;\Longrightarrow\; (x-6)^2 = 32 \;\Longrightarrow\; x = 6 \pm 4\sqrt{2} \]
The one thing that must not be forgotten is that the number is added to both sides. Adding it to the left only would change which equation you are solving.
Figure (svg): Completing the square inside an equation, showing the same number added to both sides
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 285-286 — Solve ax^2 + bx + c = 0
Picture it
Example 3: an equation that neither factoring nor taking roots could touch.
Figure (svg): Completing the square inside an equation, showing the same number added to both sides
The highlighted line is the whole method. Everything above it is rearrangement and everything below it is Lesson 4.5.
Worked example
Example 3, with the book's two checks.
\[ \text{Solve } x^2 - 12x + 4 = 0 \text{ by completing the square.} \]
Move the constant across
Why: The left side is now in the form x squared plus bx.
\[ x ^{2} - 12 x = -4 \]
Compute the number to add
Why: Half of negative 12 is negative 6, and negative 6 squared is 36.
\[ \text{add } 36 \]
Add it to both sides
Why: The left becomes a perfect square and the right becomes 32.
\[ x ^{2} - 12 x + 36 = 32 \]
Write the left as a binomial squared and take roots
Why: The bracket contains negative 6, the halved value.
\[ x - 6 = +- \sqrt{32} \]
Solve and simplify
Why: Adding 6 and simplifying the radical, since 32 is 16 times 2.
\[ x = 6 + - 4 \sqrt{2} \]
Figure (svg): The solution to Worked example leading coefficient 1 shown as a ladder of expressions, one row per algebraic move
\[ x = 6 \pm 4\sqrt{2} \]
Verify: check against a graph
Why: Four root 2 is about 5.657, so the solutions are about 11.657 and about 0.343. Graphing y equals x squared minus 12x plus 4 gives x-intercepts at about 0.34 and 11.66, matching. The two roots also average to 6, which is the vertex's x value, as the symmetry of a parabola requires.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 285-285
Ranking
Solving any quadratic by completing the square.
Put in order
Why: Dividing comes first because the halve-and-square rule is stated for a leading coefficient of 1 and gives the wrong constant otherwise. Moving the constant comes before completing so that what remains is exactly x squared plus bx, which is the expression the rule applies to.
Worked example
Example 4. Divide first, and notice where the answer ends up.
\[ \text{Solve } 2x^2 + 8x + 14 = 0 \text{ by completing the square.} \]
Divide every term by 2
Why: The equation becomes x squared plus 4x plus 7 equals zero.
\[ x ^{2} + 4 x + 7 = 0 \]
Move the constant
Why: Subtracting 7 leaves the left in the form x squared plus bx.
\[ x ^{2} + 4 x = -7 \]
Add half the middle coefficient squared
Why: Half of 4 is 2, and 2 squared is 4, added to both sides.
\[ x ^{2} + 4 x + 4 = -3 \]
Write as a binomial squared and take roots
Why: The right side is negative, so the roots are imaginary.
\[ x + 2 = +- \sqrt{-3} \]
Write in terms of i
Why: The square root of negative 3 is i root 3.
\[ x = -2 + - i \sqrt{3} \]
Figure (svg): The solution to Worked example leading coefficient not 1 shown as a ladder of expressions, one row per algebraic move
\[ x = -2 \pm i\sqrt{3} \]
Verify: substitute one solution
Why: For negative 2 plus i root 3, the bracket x plus 2 is i root 3, whose square is negative 3, and negative 3 plus 4 is 1... more directly: x squared plus 4x plus 7 evaluated there is the quantity x plus 2, squared, plus 3, which is negative 3 plus 3, or 0. It checks. Note that Lesson 4.6 is what lets this equation have solutions at all — the parabola never crosses the axis.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 286-286
Trap
\[ x^2 - 12x = -4 \]
Add 36 to complete the square on the left
Why: The number is added where it is needed and the right side is left alone.
\[ (x - 6)^2 = -4 \quad \text{(wrong)} \]
The right side is now negative, so this claims the equation has imaginary roots — but the graph crosses the axis twice, so it plainly has two real ones.
\[ x^2 - 12x + 36 = -4 + 36 \;\Longrightarrow\; (x-6)^2 = 32 \]
Add the same number to both sides
Why: Completing the square inside an equation is a balanced operation, not a rewriting of one side.
\[ x = 6 \pm 4\sqrt{2} \approx 0.34 \text{ or } 11.66 \]
A quick sanity check catches this instantly: if a right side that should be positive comes out negative, look for a missing addition. This is different from Example 4, where the right side is genuinely negative after both additions.
Fill the middle
Example 3, at the completing step.
Fill in the blanks
x^2 - 12x + 36 = -4 + 36 \;\Longrightarrow\; (x - 6)^2 = 32
Why: The same 36 goes on both sides, so the right becomes negative 4 plus 36, which is 32. Writing the addition explicitly on the right, rather than doing it in your head, is the single best guard against the most common error in this lesson.
Sorting
Look at the right side after the square is completed.
Sort into buckets
Sort each equation by the kind of roots it has.
One number decides the whole character of the answer. Lesson 4.8 gives that number a name — the discriminant — and computes it without completing the square at all.
Comparison
Fill the blanks. The only difference appears at the very end.
Comparison matrix
| Step | x^2 - 12x + 4 = 0 | 2x^2 + 8x + 14 = 0 |
|---|---|---|
| Divide first? | no, a is already 1 | yes, divide by 2 |
| Number added | 36 | 4 |
| After completing | (x - 6)^2 = 32 | (x + 2)^2 = -3 |
| Roots | 6 +- 4 sqrt(2), real | -2 +- i sqrt(3), imaginary |
The method did not adapt to the second equation; it simply reported what was there. That is what makes it general.
Section
Section 4
Concept
Building a quadratic model is unchanged from Lessons 4.3 and 4.4: write a verbal model, substitute, and put the equation in a usable form. What changes is that completing the square finishes the job whether or not the quadratic factors.
\[ 7x(x + 6) = 112 \;\Longrightarrow\; x^2 + 6x = 16 \;\Longrightarrow\; (x+3)^2 = 25 \]
Dividing by the leading coefficient early is the move that makes the arithmetic manageable, and it is legal in an equation for the same reason it was in Lesson 4.4: dividing both sides by a non-zero number preserves the solutions.
Figure (svg): A rectangle of width x plus six and length seven x, with area one hundred twelve
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 286-286 — Example 5
Picture it
Example 5: sides 7x and x plus 6, area 112.
Figure (svg): A rectangle of width x plus six and length seven x, with area one hundred twelve
The root x equal to negative 8 satisfies the equation but would give sides of negative 56 and negative 2, so it is rejected. The answer is x equal to 2, giving a 14 by 8 rectangle.
Worked example
Example 5. Area gives the equation; completing the square solves it.
\[ \text{A rectangle has sides } 7x \text{ and } x + 6 \text{ and area } 112. \text{ Find } x. \]
Write the area equation
Why: Length times width equals area.
\[ 7 x(x + 6) = 112 \]
Expand and divide by 7
Why: Distributing gives 7x squared plus 42x, and dividing every term by 7 makes the leading coefficient 1.
\[ x ^{2} + 6 x = 16 \]
Complete the square
Why: Half of 6 is 3, and 3 squared is 9, added to both sides.
\[ x ^{2} + 6 x + 9 = 25 \]
Take roots and solve
Why: The left is x plus 3, squared, and the root of 25 is 5.
\[ x = -3 + - 5 \]
Reject the impossible root
Why: The roots are 2 and negative 8, and negative 8 would give negative side lengths.
\[ x = 2 \]
Figure (svg): The solution to Worked example the rectangle shown as a ladder of expressions, one row per algebraic move
\[ x = 2 \;\Longrightarrow\; 14 \text{ by } 8 \]
Verify: compute the area
Why: At x equal to 2 the sides are 7 times 2, or 14, and 2 plus 6, or 8, and 14 times 8 is 112 — the given area. At x equal to negative 8 the sides would be negative 56 and negative 2, whose product is also 112, which is why the equation accepts it and the geometry does not.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 286-286
Fill the middle
Example 5, at the division.
Fill in the blanks
7x^2 + 42x = 112 \;\Longrightarrow\; x^2 + 6x = 16
Why: Every term is divided by 7, including the right side: 112 over 7 is 16. Forgetting to divide the right side is the other half of this error, and it produces an equation with the wrong solutions rather than an obviously broken one.
Worked example
Guided Practice 11 and 12. Both start as products.
\[ \text{Solve } 6x(x + 8) = 12 \text{ and } 4p(p - 2) = 100. \]
Expand and divide the first
Why: Six x squared plus 48x equals 12, and dividing by 6 gives x squared plus 8x equals 2.
\[ x ^{2} + 8 x = 2 \]
Complete and solve the first
Why: Half of 8 is 4, and 4 squared is 16, giving 18 on the right; the root of 18 is 3 root 2.
\[ x = -4 + - 3 \sqrt{2} \]
Expand and divide the second
Why: Four p squared minus 8p equals 100, and dividing by 4 gives p squared minus 2p equals 25.
\[ p ^{2} - 2 p = 25 \]
Complete and solve the second
Why: Half of negative 2 is negative 1, and its square is 1, giving 26 on the right.
\[ p = 1 + - \sqrt{26} \]
Figure (svg): The solution to Worked example two equations that do not factor shown as a ladder of expressions, one row per algebraic move
\[ x = -4 \pm 3\sqrt{2}; \qquad p = 1 \pm \sqrt{26} \]
Verify: check that factoring really would have failed
Why: For the first, standard form is 6x squared plus 48x minus 12 equals zero, or x squared plus 8x minus 2 after dividing; no integer pair multiplies to negative 2 and sums to 8. For the second, no integer pair multiplies to negative 25 and sums to negative 2. Both roots are irrational, which is exactly the case factoring cannot reach.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 286-286
Error analysis
A student solves the rectangle equation without dividing by 7 first.
Annotate
On: \( 7x^2 + 42x = 112, \text{ so add } (42/2)^2 = 441 \text{ to both sides} \)
The rule has a precondition, and the precondition is the first step of the procedure for a reason. Dividing costs one line and makes every later number smaller.
Prediction
Commit before reasoning.
Predict first
The equation gives x equal to 2 or negative 8. Why is only 2 reported?
Correct: At negative 8 the side lengths would be negative, which is impossible.
\[ 7(-8) = -56, \quad -8 + 6 = -2, \quad (-56)(-2) = 112 \]
Why: Substituting negative 8 gives sides of negative 56 and negative 2, whose product really is 112, so the equation is satisfied. What fails is the geometry: a rectangle cannot have negative sides. This is the same rejection as the field in Lesson 4.3 and the quilt in Lesson 4.4 — the model admits solutions the situation does not.
Comparison
Fill the blanks. Same setup, different finish.
Comparison matrix
| Model | Equation | Method that finishes it |
|---|---|---|
| Field, Lesson 4.3 | x^2 + 1000x - 240,000 = 0 | factoring: (x - 200)(x + 1200) |
| Quilt, Lesson 4.4 | 2x^2 + 9x - 5 = 0 | factoring: (2x - 1)(x + 5) |
| Rectangle, here | x^2 + 6x - 16 = 0 | either; it happens to factor |
| 6x(x + 8) = 12 | x^2 + 8x - 2 = 0 | completing the square only |
The setups are interchangeable and only the finish differs. That is why it is worth having both methods: try factoring, and complete the square when it fails.
Two truths and a lie
All three are about modelling with quadratics.
Eliminate the wrong options
Two of these are true. Knock those out and keep the false one.
Survives elimination: B
Why: The survivor is the false one. Most real quadratics do not factor over the integers; the ones that do are chosen that way by textbook authors. An equation that resists factoring is ordinary, and completing the square handles it without any suggestion that the setup was wrong.
Section
Section 5
Concept
Completing the square on a quadratic function produces vertex form, so the vertex can be read off directly. When the leading coefficient is not 1, factor it out of the first two terms before completing, and remember that the number added inside the bracket is multiplied by that coefficient.
\[ y = x^2 - 10x + 22 \;\Longrightarrow\; y = (x - 5)^2 - 3, \quad \text{vertex } (5, -3) \]
This closes the last gap in Lesson 4.2. That lesson could read a vertex off vertex form but could not produce it; completing the square produces it from any quadratic.
Figure (svg): The height of a hit baseball against time, with the vertex marked at three seconds and one hundred forty-seven feet
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 287-287 — Write a quadratic function in vertex form
Picture it
Example 7: height equals negative 16t squared plus 96t plus 3.
Figure (svg): The height of a hit baseball against time, with the vertex marked at three seconds and one hundred forty-seven feet
The vertex is at 3 seconds and 147 feet, so the maximum height is 147 feet. Nothing was guessed from the graph; the vertex was produced algebraically.
Worked example
Example 6. The equation is kept balanced by adding to both sides.
\[ \text{Write } y = x^2 - 10x + 22 \text{ in vertex form and give the vertex.} \]
Prepare to complete the square
Why: Set aside the constant 22 and look at x squared minus 10x.
\[ y = (x ^{2} - 10 x) + 22 \]
Find the number to add
Why: Half of negative 10 is negative 5, and its square is 25.
\[ \text{add } 25 \]
Add it to both sides
Why: Adding 25 inside the bracket means adding 25 to the left as well.
\[ y + 25 = (x ^{2} - 10 x + 25) + 22 \]
Write the bracket as a square and solve for y
Why: The bracket is x minus 5, squared, and subtracting 25 from both sides leaves 22 minus 25.
\[ y = (x - 5) ^{2} - 3 \]
Figure (svg): The solution to Worked example write in vertex form shown as a ladder of expressions, one row per algebraic move
\[ y = (x-5)^2 - 3, \qquad \text{vertex } (5, -3) \]
Verify: expand back to standard form
Why: The quantity x minus 5, squared, is x squared minus 10x plus 25, and subtracting 3 gives x squared minus 10x plus 22 — the original function. The vertex formula of Lesson 4.1 agrees too: negative b over 2a is 10 over 2, or 5, and substituting gives negative 3.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 287-287
Matching
Halve the middle coefficient; the constant adjusts to compensate.
Match the pairs
Why: In each case the number inside the bracket is half the middle coefficient and the constant outside is the original constant minus that number squared. Two of these have vertices below the axis and two do not, which tells you immediately how many x-intercepts each has.
Worked example
Example 7. Watch what happens to the number added.
\[ \text{Find the maximum of } y = -16t^2 + 96t + 3. \]
Factor the leading coefficient from the first two terms
Why: Negative 16 comes out, leaving t squared minus 6t inside.
\[ y = -16(t ^{2} - 6 t) + 3 \]
Find the number that completes the bracket
Why: Half of negative 6 is negative 3, and its square is 9.
\[ 9\text{ goes inside} \]
Add the right amount to both sides
Why: Adding 9 inside a bracket multiplied by negative 16 adds negative 144 to the right, so negative 144 must be added to the left too.
\[ y - 144 = -16(t ^{2} - 6 t + 9) + 3 \]
Write as a square and solve for y
Why: The bracket is t minus 3, squared, and adding 144 to both sides gives 147.
\[ y = -16(t - 3) ^{2} + 147 \]
Read the maximum
Why: The vertex is (3, 147) and the parabola opens down.
\[ \text{maximum } 147\text{ feet} \]
Figure (svg): The solution to Worked example a maximum, with a leading coefficient shown as a ladder of expressions, one row per algebraic move
\[ \text{vertex } (3, 147), \quad y_{\max} = 147 \text{ ft} \]
Verify: evaluate the original at t equal to 3
Why: Negative 16 times 9 is negative 144, plus 96 times 3, or 288, plus 3, gives 147 — matching. The textbook's Avoid Errors note is about exactly the third step: what is added to the left is negative 16 times 9, not 9, because the 9 sits inside a bracket that is being multiplied.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 287-287
Error analysis
A student rewrites the baseball function in vertex form.
Annotate
On: \( y + 9 = -16(t^2 - 6t + 9) + 3 \;\Longrightarrow\; y = -16(t - 3)^2 - 6 \)
Whatever goes inside the bracket is multiplied by the coefficient outside before it reaches the other side. Checking the vertex against the original function catches this in one substitution.
Fill the middle
Guided Practice 14.
Fill in the blanks
y = x^2 + 6x + 3 = (x + 3)^2 - 6
Why: Nine was added to make the perfect square, so 9 must be subtracted to leave the function unchanged: 3 minus 9 is negative 6. The vertex is therefore at negative 3 and negative 6. This adding-and-subtracting-the-same-number form is an alternative to the book's add-to-both-sides layout, and it gives the same answer.
Ranking
Rewriting a quadratic function with a leading coefficient.
Put in order
Why: Step one factors out of the first two terms only, leaving the constant outside — putting the constant inside the bracket as well is a common and expensive slip. Step three is where the coefficient reappears, and it is the one the textbook's Avoid Errors note is written about.
Prediction
Commit before reasoning.
Predict first
Lesson 4.2 read the vertex off vertex form. What has changed?
Correct: Any quadratic can now be put into vertex form.
\[ y = a(x - h)^2 + k \quad \text{for every quadratic, with } h = -\tfrac{b}{2a} \]
Why: Completing the square converts standard form into vertex form for every quadratic, whatever the leading coefficient. The vertex formula from Lesson 4.1 is still a faster route to the vertex alone, but it does not give you the whole rewritten function — and it is really just this calculation done once in general, which is exactly how Lesson 4.8 will derive the quadratic formula.
Comparison
Fill the blanks. Speed first, generality last.
Comparison matrix
| Method | Try it when | Limitation |
|---|---|---|
| Taking square roots | there is no first-power term | needs the equation to be a square equal to a number |
| Factoring | the coefficients are small integers | most quadratics do not factor |
| Completing the square | the other two fail, or you want vertex form | slower, and fractions appear for odd b |
| All three | give exact answers | none is wrong, only slower or narrower |
The order is a habit worth building: look for the missing middle term, then try factoring for ten seconds, then complete the square without regret.
Pattern
One routine, whether you are solving or rewriting.
For a function the last step is different: solve for y instead of taking roots, and read the vertex straight out of the result.
OpenStax Algebra and Trigonometry 2e, §2.5 Quadratic Equations §2.5
Check
Completing the square on an expression.
Check your understanding
What value of c makes x^2 + 16x + c a perfect square trinomial?
Answer: A
Why: Half of 16 is 8, and 8^2 is 64, giving x^2 + 16x + 64 = (x + 8)^2.
Check
Solving. Add to both sides.
Check your understanding
Solve x^2 - 12x + 4 = 0 by completing the square.
Answer: A
Why: Adding 36 to both sides gives (x - 6)^2 = 32, so x = 6 +- sqrt(32) = 6 +- 4 sqrt(2).
Check
Vertex form. Watch the leading coefficient.
Check your understanding
Find the maximum of y = -16t^2 + 96t + 3.
Answer: A
Why: Completing the square gives y = -16(t - 3)^2 + 147, so the vertex is (3, 147).
Real world
A rectangular vegetable bed is to be built against a wall, so only three sides need fencing. There are 40 feet of fencing available.
Discussion prompt
Write the enclosed area as a function of the width, put it in vertex form by completing the square, and find the dimensions giving the largest area.
Hint: If the width is w, the two sides use 2w feet and the remaining side uses 40 minus 2w.
Answer:
\[ A(w) = w(40 - 2w) = -2w^2 + 40w = -2(w^2 - 20w) \]
\[ A(w) = -2(w^2 - 20w + 100) + 200 = -2(w - 10)^2 + 200 \]
The vertex is (10, 200), so the bed should be 10 feet deep and 20 feet along the wall, enclosing 200 square feet.
Two things are worth noticing. The 100 added inside the bracket contributed negative 200 to the right, so 200 had to be added back — exactly the baseball calculation. And the optimal shape has the side along the wall twice the depth, which is not a square; the wall means one dimension is effectively free, and the optimum shifts accordingly.
Commit first
Answer, then rate your confidence honestly.
Predict first
Can every quadratic equation be solved by completing the square?
Correct: Yes — the method never fails.
\[ ax^2 + bx + c = 0 \;\Longrightarrow\; \left(x + \tfrac{b}{2a}\right)^2 = \tfrac{b^2 - 4ac}{4a^2} \]
Why: Half of b always exists, its square always exists, and taking roots always works once the right side is a number — real if it is positive, imaginary if it is negative, thanks to Lesson 4.6. The method needs no integer factors, no particular leading coefficient and no assumption about the answer. That universality is exactly why Lesson 4.8 can carry the calculation out on a general quadratic and get a formula that works every time.
Explain it
They can factor and they can take square roots, and they think that covers everything.
Discussion prompt
In four sentences or fewer, explain what completing the square is for and why the number you add is half of b, squared.
Hint: Draw the area picture in words.
Answer:
Taking square roots only works when one side is a square, and most quadratics do not arrive that way. Completing the square adds a number to both sides that forces the left side into a square, after which the old method finishes it.
The number is half of b squared because of the area picture: x squared plus bx is a square of side x with a rectangle of width b stuck to it, and if you split that rectangle into two strips of width half of b and put one underneath, you have a square of side x plus half of b with one corner missing. That corner is half of b, squared.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For both sides, write the addition explicitly on the right rather than doing it in your head. For dividing, make it step one on every problem, even when a is already 1, so it never gets skipped. For the outside coefficient, write out the multiplication before moving anything across. For radicals, run down the squares 4, 9, 16, 25, 36 and test each against the radicand. Do five of your chosen kind rather than twenty mixed ones.
Connect it up
Paper. Fifteen minutes.
Draw it
Take y equals x squared minus 12x plus 4 and work it four ways on one page. Top left: draw the area diagram for x squared minus 12x, showing the square, the two strips of width 6 and the missing corner, and label the corner's area. Top right: solve x squared minus 12x plus 4 equals zero by completing the square, one line per move, with the addition written on both sides. Bottom left: rewrite the function in vertex form and mark the vertex, then sketch the parabola with both x-intercepts labelled with their exact values. Bottom right: try to factor the same trinomial, list every integer pair whose product is 4, and write one sentence saying why the search fails even though the equation has two perfectly good real solutions. In a margin, write what number you would add for a middle coefficient of negative 9 rather than negative 12.
If your margin answer is a whole number, check it: half of negative 9 is not an integer, so the constant should be a fraction.
Recap
Five things, and the third one means you can now solve any quadratic equation at all.
| If you see | Then |
|---|---|
| A perfect square trinomial | Write it as a binomial squared at once |
| x^2 + bx with no constant | Add half of b, squared |
| A leading coefficient not 1 | Divide, or factor it out first |
| A number added inside a bracket | Multiply it by the outside coefficient |
| A negative right side after completing | The roots are imaginary |
| A request for a maximum | Complete the square and read k |
Completing the square works every time but takes five or six lines. Lesson 4.8 performs it once on a general quadratic and keeps the result as a formula, so the work never has to be repeated.
McDougal Littell Algebra 2 (Texas Edition), Ch. 4 Quadratic Functions and Factoring — Lesson 4.7 Complete the Square §4.7, pp. 284-289 — everything on these slides traces back here
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