HSN.CN.C.7Common CoreMathNumber and QuantityGrades 9-12
HSN.CN.C.7: Solving Quadratic Equations with Complex Solutions
In plain English: HSN.CN.C.7 is the Common Core number and quantity standard that asks students to solve quadratic equations with real coefficients whose solutions are not real numbers. Students take square roots of negative numbers, complete the square or use the quadratic formula, and write the two solutions as a conjugate pair a + bi and a - bi. It is usually taught in Algebra II.
Solve quadratic equations with real coefficients that have complex solutions.
Common Core State Standards for Mathematics · Domain: The Complex Number System (CN) · Cluster: Use complex numbers in polynomial identities and equations. Also written as HSN-CN.C.7 or N-CN.7 · Official standard
Students solve quadratic equations with real coefficients whose solutions are complex numbers. They start from equations that earlier courses labeled "no real solution," such as x² + 9 = 0, and use i² = -1 to find the actual solutions. They then solve harder equations by completing the square and by the quadratic formula, and write each answer in the form a + bi with the real and imaginary parts simplified separately.
Throughout the lesson, students check solutions by substitution and notice that the solutions always form a conjugate pair a + bi and a - bi. The graph of y = ax² + bx + c is used only as a signal: when it has no x-intercepts, the solutions are not real.
Learning Objectives
By the end of this lesson, students will be able to:
Rewrite the square root of a negative number as a multiple of i, for example √(-20) = 2i√5
Solve quadratic equations with real coefficients and complex solutions by taking square roots and by completing the square
Use the quadratic formula when the discriminant is negative and write both solutions in the form a + bi
Check a complex solution by substitution and explain why the two solutions are complex conjugates
Interpret complex solutions in a context as a value that the quantity never reaches
Prior Knowledge Required
Students should already be comfortable with:
The imaginary unit i with i² = -1 and the form a + bi HSN.CN.A.1
Adding, subtracting and multiplying complex numbers HSN.CN.A.2
Solving quadratic equations with real solutions by square roots, completing the square and the formula HSA.REI.B.4
Write three equations on the board and give students three minutes to decide, without a calculator, which ones have real solutions.
Warm-Up Prompt
"Solve if you can: (1) x² - 9 = 0, (2) x² + 9 = 0, (3) (x - 1)² = -4. Then compute i², (3i)² and (-3i)². What do your last three answers suggest about equation (2)?"
Equation (1) gives x = ±3. For (2), students usually say "no solution" because no real number squares to -9. The computations i² = -1, (3i)² = -9 and (-3i)² = -9 show that 3i and -3i both work. Ask students to use the same idea on (3): x - 1 must be a number whose square is -4, so x - 1 = ±2i and x = 1 ± 2i. Tell the class that today's goal is to replace "no real solution" with the actual complex solutions.
Direct Instruction20 minutes
Stress the phrase real coefficients in the standard: every equation today has real numbers a, b and c, but its solutions may not be real. Give students a five-step routine and model it on each worked example.
Write the equation in standard form ax² + bx + c = 0 and read off a, b and c with their signs.
Pick a method from the form: take square roots if there is no x-term or the squared expression is already isolated; complete the square when a = 1 and b is even; otherwise use the quadratic formula.
Rewrite the square root of a negative number using i: for k > 0, √(-k) = i√k.
Write each solution as a + bi: divide both the real part and the imaginary part by 2a and simplify each one.
Check one solution by substitution, using i² = -1, and confirm that the second solution is the conjugate of the first.
Square roots, no x-term
Solve 2x² + 50 = 0. Isolate x² first.
Equation: x² = -25, so x = 5i or x = -5i
Square roots, isolated square
Solve (x - 3)² = -16. The squared expression is already alone.
Equation: x - 3 = ±4i, so x = 3 + 4i or x = 3 - 4i
Completing the square
Solve x² - 8x + 20 = 0. Here a = 1 and b = -8 is even (Diagram 1).
Equation: (x - 4)² = -4, so x = 4 ± 2i
Quadratic formula, a ≠ 1
Solve 2x² - 2x + 5 = 0 with a = 2, b = -2, c = 5 (Diagram 2).
Solve 3x² + 7 = 2x. Rewrite it as 3x² - 2x + 7 = 0, so the discriminant is 4 - 84 = -80.
Equation: x = (2 ± 4i√5)/6 = 1/3 ± (2√5/3)i
After the examples, point out the pattern: every time the discriminant was negative, the two solutions had the same real part and opposite imaginary parts. With real coefficients this always happens, because the ± in the formula acts only on the imaginary part. Show the check for the third example: (4 + 2i)² - 8(4 + 2i) + 20 = (12 + 16i) - 32 - 16i + 20 = 0.
Guided Practice15 minutes
Pairs solve four equations. Partner A names the method and Partner B carries it out, then they switch for the next equation. Stop after each one and ask a pair to show its check.
Guided practice equations with methods and solutions
Equation
Suggested method
Solutions
x² + 12 = 0
Square roots
±2i√3
x² + 10x + 34 = 0
Completing the square
-5 ± 3i
4x² + 4x + 17 = 0
Quadratic formula
-1/2 ± 2i
x² - x + 1 = 0
Quadratic formula
1/2 ± (√3/2)i
Listen for three errors: writing √(-12) as -2√3, dividing only the real part by 2a (for the third equation, (-4 ± 16i)/8 is -1/2 ± 2i, not -1/2 ± 16i), and stopping at "no real solutions" without finishing.
Independent Practice10-15 minutes
Students solve six equations on their own and write every answer in a + bi form: 5x² + 45 = 0 (±3i), (x + 2)² + 9 = 0 (-2 ± 3i), x² - 6x + 10 = 0 (3 ± i), 2x² + 4x + 7 = 0 (-1 ± (√10/2)i), x² + 2x = -5 (-1 ± 2i) and 9x² - 12x + 5 = 0 (2/3 ± (1/3)i). Early finishers check two of their answers by substitution and graph y = 9x² - 12x + 5 to confirm that it has no x-intercepts.
Closure5 minutes
Exit ticket: (1) Solve x² + 4x + 5 = 0. (Answer: -2 ± i.) (2) A quadratic equation with real coefficients has the solution 6 - i. Name the other solution and explain how you know. (Answer: 6 + i, the conjugate.) (3) In one sentence, explain why "no solution" is not a complete answer for x² + 1 = 0.
Differentiation Strategies
For Struggling Students
Give a two-column organizer: the left column holds the formula step with the radical, the right column rewrites it with i before any division
Start with equations that need only square roots, such as x² + 36 = 0, before moving to the formula
Have students circle the real part and box the imaginary part before dividing by 2a, so that both get divided
For Advanced Students
Ask students to prove that if a, b and c are real and b² - 4ac < 0, the two solutions are conjugates, using the formula
Ask for a quadratic equation with integer coefficients whose solutions are 5 ± 2i, and explain the link between the solutions and the coefficients
Compare y = x² - 6x + 13 and y = x² - 6x + 5: how does moving the parabola up change the solutions from real to complex?
Assessment Guidance
What to Look For
Check that students finish the problem instead of stopping at "no real solutions," and that they write answers as a + bi with each part simplified, such as 2/3 ± (1/3)i rather than (12 ± 6i)/18. Watch for the sign error √(-k) = -√k. Ask students to say why the second solution is the conjugate of the first; a student who can explain this has understood where the ± and the i come from.
02
Classroom Activities
3 Activities
1
Error Analysis Stations
20 minGroups of 3-4
Four stations each show a solved quadratic equation that contains one error. Groups find the error, explain it in a sentence and write a correct solution. This targets the errors that make complex answers wrong even when the method is right.
Station Cards
Station A: x² + 8x + 25 = 0, student answer x = -4 ± 3 (the i was dropped; correct: -4 ± 3i)
Station B: 3x² + 48 = 0, student answer x = ±4 (the negative sign under the root was ignored; correct: ±4i)
Station C: x² - 2x + 26 = 0, student answer x = 2 ± 10i (the formula result (2 ± 10i)/2 was not divided; correct: 1 ± 5i)
Station D: 2x² + 6x + 9 = 0, student answer x = -3/2 ± 6i (only the real part was divided by 4; correct: -3/2 ± (3/2)i)
Procedure
Groups spend 4 minutes at each station and record the error, the reason and the corrected solution on their sheet
At each station, one member checks the corrected solution by substitution before the group moves on
End with a whole-class vote: which error would be easiest to make on a test, and what habit prevents it?
Modification for Distance Learning
Post each station as a slide in a shared deck. Breakout groups annotate the slide with the error and the fix, and the teacher reviews the four slides with the class afterwards.
2
Equation and Solution Match
15 minPairs
Each pair receives 8 equation cards and 10 solution cards. Two of the solution cards are distractors built from typical errors. Pairs match each equation to its solutions and must show the work that justifies each match.
Pairs sort the equations first by method: square roots, completing the square or the formula
They solve, find the matching solution card, and staple the two cards to a record sheet with the work
For each distractor, pairs write which equation it was meant to trap and what error produces it
Challenge Variation
Pairs write two new equation cards whose solutions are 3 ± 4i and -2 ± i√5, trade them with another pair and solve each other's cards.
3
Heights That Are Never Reached
15 minPairs
Students use a projectile model to see what a complex solution means in a context. A ball is thrown upward from a height of 4 feet, and its height in feet after t seconds is h = -16t² + 48t + 4.
Procedure
Graph h and find the maximum height: 40 feet at t = 1.5 seconds
Solve h = 36 (real solutions t = 1 and t = 2) and h = 50 (complex solutions). For h = 50 the equation becomes 16t² - 48t + 46 = 0, with discriminant 2304 - 2944 = -640, so t = 3/2 ± (√10/4)i
Each pair writes one sentence explaining what the complex solutions say about the ball
Discussion Questions
For which heights does the equation h = k have real solutions? For which does it have complex solutions?
The real part of both complex solutions is 3/2. Where does 3/2 appear on the graph?
Why is a complex time not a time at which something happens?
03
Diagrams & Visual Aids
2 diagrams
Diagram 1: A Parabola with No x-Intercepts and Its Complex Solutions
The graph of y = x² - 8x + 20 is drawn to scale. Its vertex (4, 4) is above the x-axis and the parabola opens up, so the equation x² - 8x + 20 = 0 has no real solutions. Completing the square gives (x - 4)² = -4 and the complex solutions 4 ± 2i. The graph shows that the solutions are not real, but it does not show the complex solutions themselves.
Diagram 2: From a Negative Discriminant to a + bi
The five steps for 2x² - 2x + 5 = 0. The discriminant is negative, so the square root becomes 6i. Both the real part 2 and the imaginary part 6i are divided by 2a = 4, which gives the conjugate pair 1/2 ± (3/2)i.
04
Homework Assignment
~30 min
HSN.CN.C.7 Homework: Quadratic Equations with Complex Solutions
Directions: Show all work. Write every non-real solution in the form a + bi or a ± bi, with the real part and the imaginary part simplified separately. Check at least one solution in each problem by substitution.
Part 2: Completing the Square and the Formula (Problems 3-4)
Solve x² + 12x + 45 = 0 by completing the square. Show the form (x - p)² = q before you take square roots.
Use the quadratic formula to solve (a) 5x² - 4x + 1 = 0 and (b) 3x² + 5x + 4 = 0. State the discriminant of each equation first.
Part 3: Checking and Interpreting (Problems 5-6)
A rectangle has a perimeter of 20 meters. Let w be its width. Write and solve an equation to find w if the area is 30 square meters. What do your solutions tell you about such a rectangle? Find the largest possible area to support your answer.
Show by substitution that x = 3 - 2i is a solution of x² - 6x + 13 = 0. Then solve the equation with the quadratic formula and explain why the other solution is the conjugate of 3 - 2i.
Rubric
Criterion
Full Credit (2 pts)
Partial Credit (1 pt)
No Credit (0 pts)
Method
Efficient method chosen and carried out correctly
Workable method with one procedural error
No method or wrong method
Square Roots of Negatives
Every √(-k) rewritten correctly as i√k
One sign or i error
Negative radicands left or treated as positive
a + bi Form
Both solutions given, each part simplified
One solution missing or not simplified
Answers not in a + bi form
Checking and Context
Substitution check and a correct interpretation
Check or interpretation incomplete
Missing
05
Quiz: 20 Questions
Interactive, with answers
Instructions
Work through the questions in order. Your score updates as you answer, and Reset quiz clears everything so you or your students can try again.
Multiple choice: pick an option to check it. Short answer: write your answer, then reveal the model answer.
0 of 20 answered · 0 correct
Question 1 of 20 · Multiple Choice
Which expression equals √(-49)?
Answer: B
√(-49) = √49 · √(-1) = 7i. Check: (7i)² = 49i² = -49. Choice A treats the root as a negative real number, but (-7)² = 49, not -49. Choice D forgets to take the square root of 49.
Question 2 of 20 · Multiple Choice
Solve x² + 100 = 0.
Answer: D
x² = -100, so x = ±√(-100) = ±10i. Choice A ignores the negative sign, and choice B drops the solution -10i. Choice C halves 100 instead of taking its square root.
Question 3 of 20 · Multiple Choice
Solve x² = -18.
Answer: A
√(-18) = i√18 = i√(9 · 2) = 3i√2, so x = ±3i√2. Choice C loses the i, so its square is +18. Choice B halves 18 instead of taking its root.
Question 4 of 20 · Multiple Choice
Solve (x + 4)² = -25.
Answer: C
Take square roots: x + 4 = ±5i, so x = -4 ± 5i. Choice A moves the 4 with the wrong sign. Choice B treats √(-25) as 5, and choice D does not take the square root of 25.
Question 5 of 20 · Multiple Choice
The discriminant of x² + 6x + 11 = 0 is 36 - 44 = -8. What does this tell you about the solutions?
Answer: B
A negative discriminant means √(b² - 4ac) is imaginary, so the formula gives -3 ± i√2, two complex conjugate solutions. Choice D confuses "no real solutions" with "no solutions": the equation has two complex solutions.
Question 6 of 20 · Multiple Choice
Solve x² - 2x + 10 = 0.
Answer: B
x = (2 ± √(4 - 40))/2 = (2 ± √(-36))/2 = (2 ± 6i)/2 = 1 ± 3i. Choice A uses +b instead of -b. Choice C forgets to divide by 2a = 2, and choice D puts 36 in the answer without taking its square root.
Question 7 of 20 · Multiple Choice
Solve x² + 14x + 53 = 0 by completing the square.
Answer: A
x² + 14x = -53. Add 49: (x + 7)² = -4, so x + 7 = ±2i and x = -7 ± 2i. Choice B has the wrong sign for the real part, choice C forgets to take the root of 4, and choice D drops the i.
Question 8 of 20 · Multiple Choice
Solve 2x² + 12x + 26 = 0.
Answer: D
Divide by 2: x² + 6x + 13 = 0. Then (x + 3)² = -4, so x = -3 ± 2i. Choice A does not divide the real part by 2. Choice C drops the i: a square cannot equal -4 in the real numbers.
Question 9 of 20 · Multiple Choice
Solve 4x² - 4x + 5 = 0.
Answer: B
b² - 4ac = 16 - 80 = -64, so x = (4 ± 8i)/8 = 1/2 ± i. Choice A divides only the real part by 8. Choice C divides by 4 instead of 2a = 8, and choice D uses +b instead of -b.
Question 10 of 20 · Multiple Choice
A quadratic equation with real coefficients has the solution 7 + 2i. What is its other solution?
Answer: C
With real coefficients, the formula gives -b/(2a) ± (imaginary part), so the solutions are conjugates: 7 + 2i and 7 - 2i. Choice A is the negative of the solution, not its conjugate: the real part stays the same and only the sign of the imaginary part changes.
Question 11 of 20 · Multiple Choice
Which equation has two non-real complex solutions?
Answer: D
For 2x² + 3x + 5 = 0, b² - 4ac = 9 - 40 = -31 < 0. The others have discriminants 1, 0 and 37, which are not negative, so their solutions are real. Choice C can look non-real because it does not factor, but a positive discriminant gives irrational real solutions.
Question 12 of 20 · Multiple Choice
A student solves x² + 6x + 25 = 0 and writes x = (-6 ± √(-64))/2 = -3 ± 8i. What went wrong?
Answer: B
√(-64) = 8i, and both parts of (-6 ± 8i) must be divided by 2, giving -3 ± 4i. Check: (-3 + 4i)² + 6(-3 + 4i) + 25 = (-7 - 24i) - 18 + 24i + 25 = 0. Choice A is wrong because 36 - 100 = -64.
Question 13 of 20 · Multiple Choice
Completing the square on x² - 10x + 34 = 0 gives which equation?
Answer: A
x² - 10x = -34. Add 25 to both sides: (x - 5)² = -9, which gives x = 5 ± 3i. Choice B has the wrong sign inside the square, and choice C gets the sign of the right side wrong: -34 + 25 = -9, not 9.
Question 14 of 20 · Multiple Choice
A ball's height in feet after t seconds is h = -16t² + 32t + 6. Solving h = 30 gives t = 1 ± (√2/2)i. What does this mean?
Answer: C
Complex solutions mean there is no real time t with h = 30. The maximum height is h(1) = 22 feet, so 30 feet is never reached. Choice A reads the imaginary part as a real number of seconds. Choice D confuses the real part with a landing time.
Question 15 of 20 · Short Answer
Solve 3x² + 2 = 0. Write the solutions in simplest form.
x² = -2/3, so x = ±i√(2/3) = ±i√6/3. x = (√6/3)i or x = -(√6/3)i. Check: ((√6/3)i)² = (6/9)(-1) = -2/3, and 3(-2/3) + 2 = 0.
Question 16 of 20 · Short Answer
Solve x² + 10x + 41 = 0.
Discriminant: 100 - 164 = -64. x = (-10 ± 8i)/2 = -5 ± 4i. Completing the square gives the same answer: (x + 5)² = -16.
Question 17 of 20 · Short Answer
Solve 2x² - 6x + 5 = 0 and write the solutions in a + bi form.
b² - 4ac = 36 - 40 = -4, so x = (6 ± 2i)/4 = 3/2 ± (1/2)i. Both the 6 and the 2i must be divided by 4.
Question 18 of 20 · Short Answer
Solve x² + x + 2 = 0.
b² - 4ac = 1 - 8 = -7, so x = (-1 ± i√7)/2 = -1/2 ± (√7/2)i.
Question 19 of 20 · Short Answer
Show by substitution that x = 1 - 4i is a solution of x² - 2x + 17 = 0.
(1 - 4i)² = 1 - 8i + 16i² = -15 - 8i. Then -2(1 - 4i) = -2 + 8i. Adding: (-15 - 8i) + (-2 + 8i) + 17 = 0, so 1 - 4i is a solution. Its conjugate 1 + 4i is the other solution.
Question 20 of 20 · Short Answer
Solve 5x² - 2x + 2 = 0.
b² - 4ac = 4 - 40 = -36, so x = (2 ± 6i)/10 = 1/5 ± (3/5)i.
0 of 20 answered · 0 correct
06
Frequently Asked Questions
10 Questions
What does HSN.CN.C.7 mean?
HSN.CN.C.7 means students can solve a quadratic equation with real coefficients even when its solutions are not real numbers. Instead of stopping at "no real solution," students find the two complex solutions, such as x = ±6i for x² + 36 = 0, and write them in the form a + bi.
Is HSN.CN.C.7 Algebra 1 or Algebra 2?
It is usually taught in Algebra II. Algebra I courses solve quadratic equations with real solutions and often stop at "no real solutions" when the discriminant is negative. Algebra II introduces i and the complex numbers (HSN.CN.A.1 and HSN.CN.A.2), and then this standard asks students to finish those equations.
How is HSN.CN.C.7 different from HSA.REI.B.4?
HSA.REI.B.4 covers solving quadratic equations in general, by every method, and includes recognizing complex solutions. HSN.CN.C.7 focuses only on equations with real coefficients whose solutions are complex, and it sits in the complex number domain. In practice the two are often taught together in Algebra II.
Why does the standard say "real coefficients"?
Because real coefficients guarantee that non-real solutions come in conjugate pairs, a + bi and a - bi. With real a, b and c, the quadratic formula gives -b/(2a) ± √(b² - 4ac)/(2a), and when b² - 4ac is negative only the imaginary part changes sign. Equations with non-real coefficients, such as x² - 2ix - 1 = 0, do not have to follow this pattern and are not part of this standard.
How do I simplify the square root of a negative number?
Take out √(-1) = i first, then simplify the positive radical. For example, √(-20) = i√20 = 2i√5 and √(-72) = 6i√2. Write the i before applying rules such as √a · √b = √(ab), because that rule fails when both numbers are negative: √(-4) · √(-9) = (2i)(3i) = -6, not √36 = 6.
What are common mistakes when solving quadratics with complex solutions?
Three errors come up often: rewriting √(-k) as -√k, dividing only the real part by 2a (turning (-2 ± 6i)/2 into -1 ± 6i instead of -1 ± 3i), and writing "no solution" when the discriminant is negative. A quick substitution check catches all three.
What do complex solutions look like on a graph?
They do not appear on the graph of y = ax² + bx + c at all: the parabola has no x-intercepts. That tells students the solutions are not real. For a leading coefficient of 1, the real part of the solutions equals the x-coordinate of the vertex. For example, y = x² - 2x + 2 has its vertex at (1, 1) and the solutions of x² - 2x + 2 = 0 are 1 ± i.
Can a quadratic equation have one real solution and one complex solution?
Not when the coefficients are real. The solutions are (-b ± √(b² - 4ac))/(2a), so they are either both real (when b² - 4ac ≥ 0) or a conjugate pair of non-real numbers with the same real part -b/(2a) (when b² - 4ac < 0). A real number is also a complex number with imaginary part 0, but students should describe the solutions as "two real" or "two non-real complex."
How should students check a complex solution?
Substitute it into the original equation and simplify with i² = -1. For x = 2 + i in x² - 4x + 5 = 0: (2 + i)² = 3 + 4i, and 3 + 4i - 8 - 4i + 5 = 0. Checking one solution is usually enough if the other one is its conjugate.
What do complex solutions mean in a word problem?
They mean the situation described by the equation cannot happen. If a height, area or price model gives complex solutions, the quantity never reaches the value in the equation. Students should say this in words and, when possible, support it with the maximum or minimum of the function.
07
Related Standards
5 standards
These standards connect to HSN.CN.C.7: prerequisites to review first, parallel standards at the same level, and next steps that build on it.
Before this lesson
HSN.CN.A.1Prerequisite
Know that i² = -1 and every complex number has the form a + bi with a and b real
