Algebra 2 · Unit 4 · Lesson 17
Completing the Square
and Complex Solutions
Learning Target
I can find complex solutions to quadratic equations by completing the square.
🧑‍🏫 Materials: Graphing technology (one device per 2–3 students) for Activity 3. Pacing: Warm-up 5 min · Activity 1 10 min · Activity 2 15 min · Activity 3 10 min · Lesson Synthesis 5 min · Cool-Down 5 min. Standards: HSA-REI.B.4.b, HSN-CN.C.7, HSA-REI.D.11. Note: This lesson builds directly on Lesson 16 — students revisit \(x^2 - 6x + 7 = 0\) (real solutions) alongside the new \(x^2 - 6x + 25 = 0\) (complex solutions). The parallel structure is intentional and should be made explicit during Activity 2 synthesis.
Warm-Up
5 min
Scrambled Quadratic Equations
Standard Form Factored Form Solutions
\(x^2 - 25 = 0\) \((x - 5i)(x + 5i) = 0\) \(\sqrt{(5)},\; -\sqrt{(5)}\)
\(x^2 - 5 = 0\) \((x - 5)(x + 5) = 0\) \(5i,\; -5i\)
\(x^2 + 25 = 0\) \((x - \sqrt{(5)})(x + \sqrt{(5)}) = 0\) \(5,\; -5\)
click to reveal factored form ▶
🧑‍🏫 Purpose: Connect quadratic equations in standard form, factored form, and their solutions — across real, irrational, and pure imaginary solution types. Correct matches: \(x^2-25=0\) ↔ \((x-5)(x+5)=0\) ↔ 5, −5 · \(x^2-5=0\) ↔ \((x-\sqrt{(5)})(x+\sqrt{(5)})=0\) ↔ \(\sqrt{(5)}\), \(-\sqrt{(5)}\) · \(x^2+25=0\) ↔ \((x-5i)(x+5i)=0\) ↔ 5i, −5i. Give 3 minutes of individual think time, then 1–2 minutes of partner comparison.
Warm-Up Synthesis
Solutions can now include complex numbers.
“How did you match the equations to their solutions? What told you which solution type to expect?”
Sample Responses click to reveal ▶
Standard Form Factored Form Solutions
\(x^2 - 25 = 0\) \((x-5)(x+5) = 0\) 5, −5
\(x^2 - 5 = 0\) \((x-\sqrt{(5)})(x+\sqrt{(5)}) = 0\) \(\sqrt{(5)}\), \(-\sqrt{(5)}\)
\(x^2 + 25 = 0\) \((x-5i)(x+5i) = 0\) 5i, −5i
Try Saying “I matched ___ to ___ because ___.”
Try Saying “The solutions are/are not real numbers because ___.”
🧑‍🏫 Key point to surface: the structure of the equation signals the solution type. For \(x^2 - 25 = 0\) — difference of squares, integer solutions. For \(x^2 - 5 = 0\) — difference of squares, irrational solutions (\(\sqrt{(5)}\)). For \(x^2 + 25 = 0\) — sum of squares, pure imaginary solutions (\(5i\)). Bridge: "In the warm-up you saw the answers. Today we will figure out how to get there."
Activity 1 · Launch
10 min total
Sometimes the Solutions Are Not Real Numbers
“What is similar about these equations? What is different?”
\((x-5)^2 = 0\)
\((x-5)^2 = 1\)
\((x-5)^2 = -1\)
The three equations all use the same expression \((x - 5)^2\). The only difference is the number on the right side.
“What do you expect will be different about the solutions?”
click to reveal ▶
🧑‍🏫 Arrange students in groups of 2. The three equations are carefully chosen so the only variable is the right side: 0, positive, and negative. Do not pre-explain — let students discover that the third equation requires imaginary numbers. Key question to pose after work time: "What is the same and what is different across the three problems, and how does that affect the solutions?"
Activity 1 · Work Time
Find All Complex Solutions — Check by Substituting
1
\((x - 5)^2 = 0\)
\(x - 5 = 0 \;\Rightarrow\; x = 5\)   Check: \((5-5)^2 = 0^2 = 0\) ✓
2
\((x - 5)^2 = 1\)
\(x - 5 = \pm 1 \;\Rightarrow\; x = 6\) or \(x = 4\)   Check: \((6-5)^2 = 1\) ✓   \((4-5)^2 = (-1)^2 = 1\) ✓
3
\((x - 5)^2 = -1\)
\(x - 5 = \pm\sqrt{(-1)} = \pm i \;\Rightarrow\; x = 5 + i\) or \(x = 5 - i\)   Check: \((5+i-5)^2 = i^2 = -1\) ✓   \((5-i-5)^2 = (-i)^2 = -1\) ✓
Try Saying “The solutions to Problem ___ are ___ because ___.”
“Problem ___ is different from Problem ___ because ___.”
🧑‍🏫 Do not show the answer key (prob-work is teacher-mode only). Look for: students who take square roots correctly with ±, students who recognize that \(\sqrt{(-1)} = i\), and students who check by substitution. Common error: writing \(x = 5 \pm 1i\) instead of \(x = 5 \pm i\) (the coefficient 1 is implied). Select one student to share Problem 3 work for the synthesis.
Activity 1 · Synthesis
The right side tells you what kind of solutions to expect.
“What is the same and what is different across the three equations? How do those similarities and differences affect the solutions?”
Right side = 0
\((x-5)^2 = 0\)
Only 0 squares to make 0.
One real solution: \(x = 5\)
Right side > 0
\((x-5)^2 = 1\)
Two real numbers square to 1.
Two real solutions: \(x = 4\) or \(x = 6\)
Right side < 0
\((x-5)^2 = -1\)
Two imaginary numbers square to −1.
Two complex solutions: \(x = 5 \pm i\)
click to reveal ▶
Try Saying “I know there are ___ real solutions because ___.”
Try Saying “The solutions are/are not real numbers because ___.”
🧑‍🏫 Emphasize the structural insight: the same process (taking the square root of both sides) works in all three cases. The only difference is what you take the square root of. When the right side is negative, the square roots are imaginary — but they still exist. This is the conceptual bridge to the upcoming MASL slides and Activity 2. If time allows: "Every negative number has two imaginary square roots. The square roots of −1 are \(i\) and \(-i\)."
MASL · Language & Notation
The Completed-Square Form
Math · We Say · Meaning
★ Math (given)
\((x + p)^2 = q\)
▲ We Say
“the quantity x plus p, squared, equals q”
● Meaning
The form produced by completing the square. If \(q > 0\): two real solutions. If \(q = 0\): one real solution. If \(q < 0\): two complex solutions.
🧑‍🏫 This is the target form for completing the square — once a quadratic is in this form, the next step is always the same: take the square root of both sides. The sign of \(q\) determines the solution type. Say the phrase aloud together. Connect to Activity 1: all three equations were already in this form with \(p = -5\) and \(q = 0, 1, -1\).
Exit Ticket · HW
5 min
Reference · How To
Find All Solutions
Reference · How To
Solving \((x + p)^2 = q\) 1. Take \(\pm\sqrt{(\ )}\) of both sides
   \(x + p = \pm\sqrt{(q)}\)
2. Isolate \(x\): subtract \(p\)
   \(x = -p \pm\sqrt{(q)}\)
Sign of \(q\):
\(q > 0\) → two real solutions
\(q = 0\) → one real solution
\(q < 0\) → two complex solutions
   \(\sqrt{(-k)} = i\sqrt{(k)}\)
Exit Ticket
On your own — find all solutions
1
\((x - 5)^2 = 2\)
2
\((x - 5)^2 = 0\)
3
\((x - 5)^2 = -2\)
🧑‍🏫 What to look for:
Problem 1: Does the student write ± before √(2)? Do they add 5 (not just write ±√(2))? Irrational solutions are acceptable unsimplified.
Problem 2: One solution only — x = 5. Watch for x = 0 (forgot the shift) or two solutions (forgot that ±0 = 0).
Problem 3: Does the student write i correctly? Common error: writing ±√(2) without the i (treating the negative as positive). Check that they add 5 to get 5 ± i√(2).
Pacing: Leave the Reference panel visible during work time.
Amplify · Activity Builder · Exit Ticket

Setup (repeat for each problem on the same screen):

1. Add a Text component with the direction and equation.

2. Add a Math Response component → name it answer1 (Problem 1), answer2 (Problem 2), answer3 (Problem 3).

3. Add a Note → click </> → paste the CL below for that problem.

Problem 1: (x − 5)² = 2
content: when answer1.submitted and answer1.latex = "5\pm\sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5 \pm \sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5\pm\sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5 \pm \sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "x=5\pm\sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "x=5 \pm \sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "x=5\pm\sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "x=5 \pm \sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5+\sqrt{(2)},5-\sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5-\sqrt{(2)},5+\sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5+\sqrt{2},5-\sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5-\sqrt{2},5+\sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5+\sqrt{(2)};5-\sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5-\sqrt{(2)};5+\sqrt{(2)}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5+\sqrt{2};5-\sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5-\sqrt{2};5+\sqrt{2}" "Correct! x = 5 ± √(2)" when answer1.submitted and answer1.latex = "5+\sqrt{(2)}" "That is one solution. What is the other? (Enter both: x = 5 + √(2), x = 5 − √(2))" when answer1.submitted and answer1.latex = "5-\sqrt{(2)}" "That is one solution. What is the other? (Enter both: x = 5 + √(2), x = 5 − √(2))" when answer1.submitted and answer1.latex = "5+\sqrt{2}" "That is one solution. What is the other? (Enter both: x = 5 + √(2), x = 5 − √(2))" when answer1.submitted and answer1.latex = "5-\sqrt{2}" "That is one solution. What is the other? (Enter both: x = 5 + √(2), x = 5 − √(2))" when answer1.submitted "Take the square root of both sides: x − 5 = ±√(2). Then add 5 to both sides. What do you get? (Enter both solutions.)" otherwise ""
Problem 2: (x − 5)² = 0
content: when answer2.submitted and answer2.numericValue = 5 "Correct! x = 5 is the only solution — only 0 squares to make 0." when answer2.submitted and answer2.latex = "x=5" "Correct! x = 5 is the only solution — only 0 squares to make 0." when answer2.submitted and answer2.numericValue = 0 "Almost — check the isolation step. You found x − 5 = 0, so x = ?" when answer2.submitted and answer2.latex = "0" "Almost — check the isolation step. You found x − 5 = 0, so x = ?" when answer2.submitted "Take the square root of both sides: x − 5 = ±0 = 0. Isolate x. What do you get?" otherwise ""
Problem 3: (x − 5)² = −2
content: when answer3.submitted and answer3.latex = "5\pm i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5 \pm i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5\pm i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5 \pm i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5\pm\sqrt{(2)}i" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5 \pm \sqrt{(2)}i" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "x=5\pm i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "x=5 \pm i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "x=5\pm i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "x=5 \pm i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5+i\sqrt{(2)},5-i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5-i\sqrt{(2)},5+i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5+i\sqrt{2},5-i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5-i\sqrt{2},5+i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5+i\sqrt{(2)};5-i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5-i\sqrt{(2)};5+i\sqrt{(2)}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5+i\sqrt{2};5-i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5-i\sqrt{2};5+i\sqrt{2}" "Correct! x = 5 ± i√(2)" when answer3.submitted and answer3.latex = "5+i\sqrt{(2)}" "That is one solution. What is the other? (Enter both: x = 5 + i√(2), x = 5 − i√(2))" when answer3.submitted and answer3.latex = "5-i\sqrt{(2)}" "That is one solution. What is the other? (Enter both: x = 5 + i√(2), x = 5 − i√(2))" when answer3.submitted and answer3.latex = "5\pm\sqrt{(-2)}" "Good setup — now simplify: √(-2) = i√(2). Rewrite your answer using i." when answer3.submitted and answer3.latex = "5 \pm \sqrt{(-2)}" "Good setup — now simplify: √(-2) = i√(2). Rewrite your answer using i." when answer3.submitted and answer3.latex = "5\pm\sqrt{(2)}" "Almost! Check the sign inside the square root. The right side is −2, so √(−2) = i√(2), not √(2)." when answer3.submitted and answer3.latex = "5 \pm \sqrt{(2)}" "Almost! Check the sign inside the square root. The right side is −2, so √(−2) = i√(2), not √(2)." when answer3.submitted "Take the square root of both sides: x − 5 = ±√(−2) = ±i√(2). Then add 5. What do you get? (Enter both solutions.)" otherwise ""
Reference · How To
Completing the Square with Complex Solutions
Step What to do Example: \(x^2 - 6x + 25 = 0\)
1 Move the constant to the right side \(x^2 - 6x = -25\)
2 Take half the coefficient of \(x\) and square it Half of \(-6\) is \(-3\).   \((-3)^2 = 9\)
3 Add that square to both sides \(x^2 - 6x + 9 = -25 + 9 = -16\)
4 Write the left side as a squared binomial \((x - 3)^2 = -16\)
5 Take the square root of both sides — use \(\pm\) \(x - 3 = \pm\sqrt{(-16)} = \pm\sqrt{(16)(-1)}\)\(= \pm\sqrt{(4)^2(-1)} = \pm 4\sqrt{(-1)} = \pm 4i\)
6 Isolate \(x\) — add 3 to both sides \(x - 3 + 3 = \pm 4i + 3\)\(x = 3 \pm 4i\)
★ When the right side is negative, use \(i\):   \(\sqrt{(-k)} = i\sqrt{(k)}\) for \(k > 0\) \(\sqrt{(-16)} = \sqrt{(16)} \cdot i = 4i\)    The solutions are complex (imaginary).
🧑‍🏫 Leave this slide on screen during Activity 2 work time. Walk through it one click at a time. Step 3 is where most errors happen — students must add the same value to both sides. Step 5 is the new extension: when the right side is negative, the square roots are imaginary. The ★ row connects to the previous lesson's imaginary number notation.
MASL · Language & Notation
Imaginary Solutions from Square Roots
Math · We Say · Meaning
★ Math (given)
\[\begin{aligned} &\text{if } (x+p)^2 = -q \\ &\text{then } x = -p \pm i\sqrt{(q)} \end{aligned}\]
▲ We Say
“If a squared binomial equals a negative number, then x equals negative p, plus or minus i times the square root of q”
● Meaning
When completing the square gives a negative right side, the solutions involve \(i\). The two solutions form a conjugate pair — same real part, opposite imaginary parts.
🧑‍🏫 This MASL slide shows the full solution form when completing the square produces a negative right side. Read it aloud together. Example: \((x-3)^2 = -16\) gives \(x = 3 \pm 4i\). The conjugate pair structure is important — it means if \(3 + 4i\) is a solution, then \(3 - 4i\) must be too. Students will see this again in Activity 3 when they connect to graphs.
Activity 2 · Work Time
15 min
Reference · How To
Finding Complex Solutions
Reference · How To
Completing the Square
\(ax^2 + bx + c = 0\)
1. Move constant right: \(x^2 + bx = -c\)
2. Compute \(\left(\dfrac{b}{2}\right)^2\), add to both sides.
3. Factor: \(\left(x + \dfrac{b}{2}\right)^2 = k\)
4. Take \(\pm\) square root.
5. If \(k < 0\): write \(\sqrt{(|k|)} \cdot i\)
6. Isolate \(x\).
Activity 2 · Work Time
Solve each equation by completing the square to find all solutions.
1
\(x^2 - 8x + 13 = 0\)
Step 1: \(x^2 - 8x = -13\)
Step 2: \(\left(\frac{-8}{2}\right)^2 = 16\)
Step 3: \(x^2 - 8x + 16 = -13 + 16\)
Step 4: \((x-4)^2 = 3\)
Step 5: \(x - 4 = \pm\sqrt{(3)}\)
Step 6: \(x = 4 \pm \sqrt{(3)}\)
2
\(x^2 - 8x + 19 = 0\)
Step 1: \(x^2 - 8x = -19\)
Step 2: \(\left(\frac{-8}{2}\right)^2 = 16\)
Step 3: \(x^2 - 8x + 16 = -19 + 16\)
Step 4: \((x-4)^2 = -3\)
Step 5: \(x - 4 = \pm\sqrt{(-3)} = \pm i\sqrt{(3)}\)
Step 6: \(x = 4 \pm i\sqrt{(3)}\)
Try Saying “The solutions to Problem 1 are ___ because the right side after completing the square is ___.”
“The two equations have the same/different solution types because ___.”
🧑‍🏫 The two equations are intentionally parallel — both add 16 to complete the square, both produce \((x-4)^2 = k\). The difference is \(k = 3\) (real solutions) vs \(k = -3\) (complex solutions). Look for students who recognize the same steps produce different solution types based only on the sign of \(k\). Common error (Problem 2): writing \(\pm\sqrt{(3)}\) without the \(i\) — treating the negative as positive. Common error (both): forgetting to add 16 to the right side too. Select one student per problem to share for synthesis.
Amplify · Activity Builder · Type-in

Setup (do once):

1. Add a Text component → include the direction "Solve by completing the square" and both equations.

2. Add two Math Response components → name them answer1 and answer2.

3. Add a Note component below each → click </> → paste the matching CL block.

4. Straight quotes only — curly quotes break CL.

Problem 1: x² − 8x + 13 = 0 (solutions: 4 ± √(3))
content: when answer1.submitted and answer1.latex = "4\pm\sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4 \pm \sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4\pm\sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4 \pm \sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "x=4\pm\sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "x=4 \pm \sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "x=4\pm\sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "x=4 \pm \sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4+\sqrt{(3)},4-\sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4-\sqrt{(3)},4+\sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4+\sqrt{3},4-\sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4-\sqrt{3},4+\sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4+\sqrt{(3)};4-\sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4-\sqrt{(3)};4+\sqrt{(3)}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4+\sqrt{3};4-\sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4-\sqrt{3};4+\sqrt{3}" "Correct! x = 4 + sqrt(3) or x = 4 - sqrt(3)." when answer1.submitted and answer1.latex = "4+\sqrt{(3)}" "That is one solution. What is the other? (Enter both: x = 4 + sqrt(3), x = 4 - sqrt(3))" when answer1.submitted and answer1.latex = "4-\sqrt{(3)}" "That is one solution. What is the other? (Enter both: x = 4 + sqrt(3), x = 4 - sqrt(3))" when answer1.submitted and answer1.latex = "4+\sqrt{3}" "That is one solution. What is the other? (Enter both: x = 4 + sqrt(3), x = 4 - sqrt(3))" when answer1.submitted and answer1.latex = "4-\sqrt{3}" "That is one solution. What is the other? (Enter both: x = 4 + sqrt(3), x = 4 - sqrt(3))" when answer1.submitted "Complete the square: add 16 to both sides. (x - 4)^2 = 3. Then x - 4 = ±sqrt(3). (Enter both: x = 4 + sqrt(3), x = 4 - sqrt(3))" otherwise ""
Problem 2: x² − 8x + 19 = 0 (solutions: 4 ± i√(3))
content: when answer2.submitted and answer2.latex = "4\pm i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4 \pm i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4\pm i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4 \pm i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4\pm\sqrt{(3)}i" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4 \pm \sqrt{(3)}i" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "x=4\pm i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "x=4 \pm i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "x=4\pm i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "x=4 \pm i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4+i\sqrt{(3)},4-i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4-i\sqrt{(3)},4+i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4+i\sqrt{3},4-i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4-i\sqrt{3},4+i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4+i\sqrt{(3)};4-i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4-i\sqrt{(3)};4+i\sqrt{(3)}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4+i\sqrt{3};4-i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4-i\sqrt{3};4+i\sqrt{3}" "Correct! x = 4 + i*sqrt(3) or x = 4 - i*sqrt(3)." when answer2.submitted and answer2.latex = "4+i\sqrt{(3)}" "That is one solution. What is the other? (Enter both: x = 4 + i*sqrt(3), x = 4 - i*sqrt(3))" when answer2.submitted and answer2.latex = "4-i\sqrt{(3)}" "That is one solution. What is the other? (Enter both: x = 4 + i*sqrt(3), x = 4 - i*sqrt(3))" when answer2.submitted and answer2.latex = "4\pm\sqrt{(3)}" "Almost! Check the sign under the square root — the right side is -3, so you need i. Rewrite sqrt(-3) = i*sqrt(3)." when answer2.submitted and answer2.latex = "4 \pm \sqrt{(3)}" "Almost! Check the sign under the square root — the right side is -3, so you need i. Rewrite sqrt(-3) = i*sqrt(3)." when answer2.submitted and answer2.latex = "4\pm\sqrt{(-3)}" "Good setup — now simplify: sqrt(-3) = i*sqrt(3). Rewrite your answer using i." when answer2.submitted and answer2.latex = "4 \pm \sqrt{(-3)}" "Good setup — now simplify: sqrt(-3) = i*sqrt(3). Rewrite your answer using i." when answer2.submitted "Complete the square: add 16 to both sides. (x - 4)^2 = -3. Then x - 4 = ±i*sqrt(3). (Enter both: x = 4 + i*sqrt(3), x = 4 - i*sqrt(3))" otherwise ""
Activity 2 · Synthesis
Same process — different sign, different solution type.
“Why does one equation have real solutions and the other have complex solutions? What is the key difference?”
Real Solutions
\(x^2 - 8x + 13 = 0\)
\((x-4)^2 = 3\)
\(x - 4 = \pm\sqrt{(3)}\)
\(x = 4 \pm \sqrt{(3)}\)
Complex Solutions
\(x^2 - 8x + 19 = 0\)
\((x-4)^2 = -3\)
\(x - 4 = \pm i\sqrt{(3)}\)
\(x = 4 \pm i\sqrt{(3)}\)
Try Saying “The solutions are real/complex because after completing the square, the right side is ___.”
Try Saying “When \(k\) is positive, the solutions are ___. When \(k\) is negative, the solutions are ___.”
🧑‍🏫 The three-equation comparison (\(k = 3, 0, -3\)) is the key structural insight of the lesson. All three equations have the form \((x-4)^2 = k\). The sign of \(k\) determines the solution type. Ask: "What equation would give \((x-4)^2 = 0\)?" (Answer: \(x^2 - 8x + 16 = 0\).) This sets up Activity 3's graphical interpretation.
Activity 3 · Work Time
10 min
Can You See the Solutions on a Graph?
Use graphing technology to graph all three functions. Then answer the questions for each equation.
Equation A
\(f(x) = x^2 - 6x + 7\)
How many real solutions does \(x^2 - 6x + 7 = 0\) have?
How many complex solutions?
2 real, 0 complex. Graph crosses x-axis in 2 places. Vertex at (3, −2) — below axis.
Equation B
\(f(x) = x^2 - 6x + 9\)
How many real solutions does \(x^2 - 6x + 9 = 0\) have?
How many complex solutions?
1 real, 0 complex. Graph touches x-axis in 1 place. Vertex at (3, 0) — on the axis.
Equation C
\(f(x) = x^2 - 6x + 25\)
How many real solutions does \(x^2 - 6x + 25 = 0\) have?
How many complex solutions?
0 real, 2 complex. Graph does not cross x-axis. Vertex at (3, 16) — above axis.
Discuss with your partner
  1. How does the graph of each function help you determine the number of real vs. complex solutions?
  2. If a graph does not cross the x-axis, does that mean the equation has no solutions? Explain.
Try Saying “The graph tells me the equation has ___ real solution(s) because ___.”
Try Saying “No x-intercepts means ___, not ___.”
🧑‍🏫 One device per 2–3 students. Students can use Desmos (desmos.com/calculator) or TI graphing calculators. The three functions all have vertex x-coordinate \(x = 3\) — emphasize this visual connection to the algebra. Key misconception to surface in synthesis: "no x-intercepts" does not mean "no solutions" — it means no real solutions, but two complex solutions still exist. See the building-on-thinking note in the activity JSON: if students plot complex solutions in the coordinate plane as if they were real, ask "What is the same and different about the points (3, 4) and (3 + 4i)?"
Activity 3 · Synthesis
No x-intercepts does not mean no solutions.
“If a graph does not cross the x-axis, what can you conclude about the solutions to the related equation?”
Crosses axis twice
\(f(x) = x^2-6x+7\)
2 real solutions
\(x = 3 \pm \sqrt{(2)}\)
Touches axis once
\(f(x) = x^2-6x+9\)
1 real solution
\(x = 3\)
Does not cross axis
\(f(x) = x^2-6x+25\)
2 complex solutions
\(x = 3 \pm 4i\)
Try Saying “If \(y = f(x)\), then \(f(x) = 0\) means ___.”
“On the graph, I see ___, so I know ___.”
Try Saying “I agree/disagree that there are no solutions because ___.”
🧑‍🏫 Key insight: complex solutions are still solutions — they just cannot be located on a graph that uses only real number lines. The graph shows whether real solutions exist and how many; it does not show complex solutions. Ask: "How many solutions does \(x^2 - 6x + 25 = 0\) have?" (Two — both complex.) "Can you see them on the graph?" (No — because complex numbers cannot be represented on a standard real-number graph.)
Lesson Synthesis
When do imaginary numbers show up?
“How do you know when imaginary numbers will appear when solving a quadratic equation by completing the square?”
Rule: Imaginary numbers appear when completing the square produces a negative number on the right side of \((x + p)^2 = q\). This happens when \(q < 0\).
Without solving: 1 real solution
\((x - 3)^2 = 5\)
Right side positive → two real solutions.
Vertex below x-axis.
Two complex solutions
\(x^2 + 2x + 5 = 0\)
Complete the square: \((x+1)^2 = -4\)
Right side negative → \(x = -1 \pm 2i\)
Try Saying “I know imaginary numbers will appear because ___.”
Try Saying “The graph of this function tells me the solutions are ___ because ___.”
🧑‍🏫 Use the discussion questions from the lesson synthesis field: (1) "How do you know when imaginary numbers will show up?" (2) "Without solving, how many solutions does \((x-3)^2 = 5\) have?" (Two real.) (3) "What about \(x^2 + 2x + 5 = 0\)?" (Two complex — vertex form is \((x+1)^2 + 4 = 0\), so \((x+1)^2 = -4\).) (4) "Is it possible for a quadratic to have exactly one complex solution?" (Yes — e.g., \((x - i)^2 = 0\) has the single solution \(x = i\). This is beyond the standard but worth naming if students ask.)
Cool-Down
5 min
Reference · How To
Make One
Reference · How To
Completing the Square 1. Move constant right.
2. Add \(\left(\dfrac{b}{2}\right)^2\) to both sides.
3. Factor: \((x + p)^2 = q\)
4. Take \(\pm\) square root.
5. If \(q < 0\): solutions involve \(i\)
6. Isolate \(x\).
Tip: Start with \((x + p)^2 = -q\) and expand. Any negative right side guarantees complex solutions.
Cool-Down
On your own
  1. Write a quadratic equation whose solutions are not real numbers.
  2. Solve your equation by completing the square. Show all steps.
🧑‍🏫 Sample responses:
Simple: \(x^2 + 9 = 0\) → \((x+0)^2 = -9\) → \(x = \pm 3i\)
Moderate: \(x^2 + 2x + 5 = 0\) → \((x+1)^2 = -4\) → \(x = -1 \pm 2i\)
Advanced: \(x^2 - 6x + 25 = 0\) → \((x-3)^2 = -16\) → \(x = 3 \pm 4i\)

What to look for: Does the student's equation actually produce a negative right side when completing the square? Are all steps shown? Is \(i\) used correctly? Is there a conjugate pair?
Common error: Writing \(x^2 - 9 = 0\) — this has real solutions 3 and −3, not complex solutions.
Amplify · Activity Builder · Open Response

Setup:

1. Add a Text component → include the full task: "Write a quadratic equation whose solutions are not real numbers. Solve your equation by completing the square. Show all steps."

2. Add a Math Response component → name it answer. (Student enters their final solutions.)

3. Add a Note component → click </> → paste CL below.

4. This is an open-ended task — CL checks for the imaginary unit in the answer and gives process feedback.

content: when answer.submitted and isBlank(answer.latex) "Enter your two solutions in the box above." when answer.submitted and answer.latex = "3i,-3i" "Correct form! Make sure your equation also produces this answer when you complete the square." when answer.submitted and answer.latex = "3i, -3i" "Correct form! Make sure your equation also produces this answer when you complete the square." when answer.submitted and answer.latex = "\pm3i" "Correct form! Make sure your equation also produces this answer when you complete the square." when answer.submitted and answer.latex = "\pm 3i" "Correct form! Make sure your equation also produces this answer when you complete the square." when answer.submitted and answer.latex = "-1+2i,-1-2i" "Correct form — a conjugate pair with real part -1. Check your equation." when answer.submitted and answer.latex = "-1+2i, -1-2i" "Correct form — a conjugate pair with real part -1. Check your equation." when answer.submitted and answer.latex = "-1\pm2i" "Correct form — a conjugate pair with real part -1. Check your equation." when answer.submitted and answer.latex = "-1 \pm 2i" "Correct form — a conjugate pair with real part -1. Check your equation." when answer.submitted and answer.latex = "3+4i,3-4i" "Correct form — a conjugate pair with real part 3. Check your equation produces (x-3)^2 = -16." when answer.submitted and answer.latex = "3+4i, 3-4i" "Correct form — a conjugate pair with real part 3. Check your equation produces (x-3)^2 = -16." when answer.submitted and answer.latex = "3\pm4i" "Correct form — a conjugate pair with real part 3. Check your equation produces (x-3)^2 = -16." when answer.submitted and answer.latex = "3 \pm 4i" "Correct form — a conjugate pair with real part 3. Check your equation produces (x-3)^2 = -16." when answer.submitted "Make sure your solutions include i — they should be complex numbers in the form a ± bi." otherwise ""