Solving Quadratics (Optional)
Learning Target
I can solve quadratic equations by completing the square or by using the quadratic formula.
🧑🏫 This lesson is optional. It revisits completing the square and the quadratic formula from an earlier course. If the pre-unit diagnostic indicates students know this material, skip this lesson. The lessons that follow require fluency with these skills, so use this lesson if students need the review. Approximate pacing: 45 minutes total.
Find the Perfect Squares
The expression \((x + n)^2\) is equivalent to \(x^2 + 2nx + n^2\).
On your own
Which expressions are equivalent to \((x + n)^2\) for some number \(n\)?
- \(x^2 + 10x + 25\)
- \(x^2 + 10x + 100\)
- \(x^2 + 7x + 12\)
- \(x^2 - 6x + 9\)
🧑🏫 Purpose: review perfect square trinomials before completing the square. Students need to recognize that the constant term must be the square of half the coefficient of \(x\). Give 2 minutes of individual think time, then cold-call 2–3 students.
The constant must be the square of half the coefficient of \(x\).
“How can you tell whether a quadratic expression is a perfect square trinomial?”
| Equation |
Reasoning |
Perfect Square? |
| \(x^2 + 10x + 25\) |
Half of 10 is 5, and \(5^2 = 25\) |
Yes |
| \(x^2 + 10x + 100\) |
Half of 10 is 5, but \(5^2 = 25 \neq 100\) |
No |
| \(x^2 + 7x + 12\) |
Half of 7 is 3.5, and \(3.5^2 = 12.25 \neq 12\) |
No |
| \(x^2 - 6x + 9\) |
Half of −6 is −3, and \((-3)^2 = 9\) |
Yes |
The binomial \(\left(x + \dfrac{b}{2}\right)^2\) expands to \(x^2 + bx + \left(\dfrac{b}{2}\right)^2\). In \(ax^2 + bx + c\) with \(a = 1\), the expression is a perfect square trinomial when \(c = \left(\dfrac{b}{2}\right)^2\).
Try Saying
“I know this is a perfect square trinomial because ___.”
“This is not a perfect square trinomial because ___.”
🧑🏫 Highlight: the key test is \(c = \left(\frac{b}{2}\right)^2\). If the constant matches, the expression is a perfect square. If it does not match, we can adjust \(c\) — that is completing the square. Bridge to today's lesson: "Today we use this idea to solve quadratic equations."
Activity 1 · Launch
15 min total
Different Ways to Solve It
Elena and Han each solved the equation \(x^2 - 6x + 7 = 0\) using a different method. Read their work. Decide whether you agree with each solution.
⏱ 1 min individual read time before partner discussion
🧑🏫 Arrange students in pairs. Read the context aloud together. Do not pre-explain \(\pm\) notation — let students encounter it in Elena and Han's work first. If a student asks, prompt: "What do you notice about what Elena wrote?" Encourage partners to explain their reasoning and reach agreement before discussing as a class.
Different Ways to Solve It
Elena — Completing the Square
\(x^2 - 6x + 7 = 0\)
Add 2 to each side: \(x^2 - 6x + 9 = 2\)
Factor: \((x - 3)^2 = 2\)
Square root: \(x - 3 = \pm\sqrt{2}\)
Solve: \(x = 3 + \sqrt{2}\) and \(x = 3 - \sqrt{2}\)
Han — Quadratic Formula
\(x^2 - 6x + 7 = 0\)
\(a = 1,\; b = -6,\; c = 7\)
\(x = \dfrac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(7)}}{2(1)}\)
\(x = \dfrac{6 \pm \sqrt{8}}{2}\)
\(x = \dfrac{6 + \sqrt{8}}{2}\) and \(x = \dfrac{6 - \sqrt{8}}{2}\)
Discuss with your partner
- Do you agree with Elena's method? Explain your reasoning.
- Do you agree with Han's method? Explain your reasoning.
- The solutions look different. Are they equivalent? How do you know?
Try Saying
“I agree with ___ because ___.”
“The solutions are/are not equivalent because ___.”
🧑🏫 Both methods are correct. The key question is #3: the solutions look different but are equivalent. Students must show that \(\sqrt{8} = 2\sqrt{2}\), so \(\frac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}\). Listen for students who simplify the radical vs. students who check by squaring. Select one of each for synthesis.
Both methods give the same solutions.
“The solutions look different. How can you show they are the same values?”
Key equivalence: \(\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}\)
So Han's solutions simplify:
\(\dfrac{6 \pm 2\sqrt{2}}{2} = 3 \pm \sqrt{2}\)
Both Elena and Han found the same two solutions: \(x = 3 + \sqrt{2}\) and \(x = 3 - \sqrt{2}\).
Elena — how she set up
Add 2 so the constant becomes \((-3)^2 = 9\).
Han — how he set up
Identify \(a = 1\), \(b = -6\), \(c = 7\) and substitute into the formula.
Try Saying
“I can show whether the two answers are equivalent by ___.”
“They are/are not the same value because ___.”
🧑🏫 Discussion questions: "How did Elena know to add 2?" (She needed the constant to be \((-3)^2 = 9\), and she already had 7, so she added 2.) "Why is there a positive 6 in Han's formula when \(b = -6\)?" (The formula has \(-b\), so \(-(-6) = 6\).) Emphasize: both strategies are valid; different-looking answers can be equivalent. Simplify radicals to check.
MASL · Language & Notation
The Plus-or-Minus Symbol
★ Math (given)
\(3 \pm \sqrt{2}\)
▲ We Say
“three plus or minus the square root of two”
● Meaning
This expression represents two values at once: \(3 + \sqrt{2}\) and \(3 - \sqrt{2}\).
🧑🏫 The \(\pm\) notation is a persistent source of confusion for multilingual learners. Emphasize: this single expression stands for two separate numbers. This is critical for interpreting quadratic formula results. Read it aloud together as a class.
MASL · Language & Notation
Completing the Square
★ Math (given)
\[\begin{aligned} x^2 + bx + \left(\frac{b}{2}\right)^2 \\ = \left(x + \frac{b}{2}\right)^2 \end{aligned}\]
▲ We Say
“completing the square”
● Meaning
Adding \(\left(\dfrac{b}{2}\right)^2\) to \(x^2 + bx\) creates a perfect square trinomial that factors as a squared binomial.
🧑🏫 Connect to the warm-up: students just tested whether expressions were perfect squares — this slide names the technique for making any quadratic into one. Emphasize: "We choose c so the expression becomes a perfect square — that's completing the square." Say the phrase aloud together.
Activity 2 · Launch
10 min total
Solve These Ones
Solve each quadratic equation with a method of your choice. Use a different method than your partner on each problem. Compare your answers — verify that they are equivalent.
⏱ 1 min — choose your first method before starting
🧑🏫 Arrange students in pairs. The key instruction: each partner uses a different method on the same problem, then they compare. This forces students to practice both strategies and verify equivalence. Circulate and identify pairs who solved the same problem with different methods for the synthesis.
Activity 2 · Work Time
Reference · How To
Solve These Ones
Reference · How To
Completing the Square
1. Half the coefficient of \(x\), then square it.
2. Add that value to both sides.
3. Factor as \(\left(x + \dfrac{b}{2}\right)^2 = k\).
4. Take \(\pm\) square root, isolate \(x\).
Quadratic Formula
\[x = \frac{-b \pm \sqrt{(b^2 - 4ac)}}{2a}\]
1. Standard form \(ax^2 + bx + c = 0\).
2. Identify \(a, b, c\) — include signs.
3. Substitute and simplify.
Activity 2 · Work Time
1 \((x - 5)^2 = 36\)
2 \((x + 3)^2 = 10\)
3 \(x^2 + 8x + 16 = 0\)
4 \(x^2 + 14x + 40 = 0\)
5 \(x^2 + 3x - 5 = 0\)
6 \(3x^2 + x - 4 = 0\)
🧑🏫 Leave the reference panel visible during work time. Look for: (1) students who recognize Problems 1–2 need no formula — just square roots, (2) students who spot Problem 3 as a perfect square with one solution, (3) students who choose different methods on Problems 4–6. Common errors: sign errors when \(b\) is negative, forgetting to simplify \(\sqrt{49} = 7\) in Problem 6.
Which method works best when?
“Why does Problem 3 have only one solution instead of two?”
Square roots
Best when there is no linear term, like \((x - 5)^2 = 36\).
Completing the square
Best when \(a = 1\) and the coefficient of \(x\) is even.
Quadratic formula
Works always. Most useful when other methods are messy.
One solution: Problem 3 is \((x + 4)^2 = 0\). The graph of \(y = x^2 + 8x + 16\) touches the \(x\)-axis at the vertex only — so there is exactly one solution.
Try Saying
“Problem 3 has ___ solution(s) because ___.”
Try Saying
“I chose ___ because ___.”
🧑🏫 Select 2–3 pairs to share: one pair who solved the same problem by different methods, and one pair who got different-looking answers and verified equivalence. Ask: "What conditions helped you choose a method?" Emphasize the three strategies and when each is efficient. If two pairs found different-looking answers for Problem 5, verify equivalence as a class.
MASL · Language & Notation
The Quadratic Formula
★ Math (given)
\[x = \frac{-b \pm \sqrt{(b^2 - 4ac)}}{2a}\]
▲ We Say
“negative b, plus or minus the square root of b squared minus 4ac, all over 2a”
● Meaning
The formula that gives the solutions to any equation in the form \(ax^2 + bx + c = 0\). Substitute the values of \(a\), \(b\), and \(c\) directly.
🧑🏫 Remind students that the quadratic formula is derived by completing the square on \(ax^2 + bx + c = 0\) — it always works. Read it aloud together. Emphasize: get to standard form first, then identify \(a\), \(b\), \(c\) including their signs before substituting.
How to solve a quadratic by completing the square
| Step |
What to do |
Example: \(x^2 - 6x + 7 = 0\) |
| 1 |
Take half the coefficient of \(x\) and square it |
Half of \(-6\) is \(-3\). \(\;(-3)^2 = 9\) |
| 2 |
Add or subtract to make the constant equal that square |
Need 9, have 7 → add 2 to both sides: \(x^2 - 6x + 9 = 2\) |
| 3 |
Write the left side as a squared binomial |
\((x - 3)^2 = 2\) |
| 4 |
Take the square root of both sides — use \(\pm\) |
\(x - 3 = \pm\sqrt{2}\) |
| 5 |
Isolate \(x\) |
\(x = 3 \pm \sqrt{2}\) |
🧑🏫 Leave this slide on screen during Activity 2 work time and during the cool-down. Walk through it one click at a time. Step 2 is where most errors happen — students forget to add the same value to both sides.
How to solve a quadratic using the quadratic formula
| Step |
What to do |
Example: \(3x^2 + x - 4 = 0\) |
| 1 |
Write in standard form \(ax^2 + bx + c = 0\) |
Already in standard form: \(a = 3,\; b = 1,\; c = -4\) |
| 2 |
Identify \(a\), \(b\), and \(c\) — include signs |
\(a = 3\), \(b = 1\), \(c = -4\) |
| 3 |
Substitute into \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) |
\(x = \frac{-(1) \pm \sqrt{(1)^2 - 4(3)(-4)}}{2(3)}\) |
| 4 |
Simplify under the radical, then the fraction |
\(x = \frac{-1 \pm \sqrt{1 + 48}}{6} = \frac{-1 \pm 7}{6}\) |
| 5 |
Evaluate the two solutions separately |
\(x = \frac{-1 + 7}{6} = 1\) or \(x = \frac{-1 - 7}{6} = -\frac{4}{3}\) |
| ★ |
Watch signs: \(-b\) means "change the sign of \(b\)" |
If \(b = -6\), then \(-b = -(-6) = 6\). Do not write \(-6\) again. |
🧑🏫 Leave this slide on screen during the cool-down. The ★ row addresses the most common error: mishandling the sign of \(b\). Walk through it one click at a time. Emphasize Step 2: include the sign of each coefficient.
Three strategies — choose wisely.
“What conditions about a quadratic equation help you decide which strategy to use?”
Reasoning
Get the equation into the form \(\text{(expression)}^2 = k\).
Completing the Square
Get the equation into the form \(x^2 + bx + \text{___} = k\).
Quadratic Formula
Write in the form \(ax^2 + bx + c = 0\).
Different methods can give answers that look different but are equivalent. Simplify radicals and fractions to verify.
Try Saying
“I would use ___ when ___.”
Try Saying
“I can check whether they are the same by ___.”
“If one person got \(5 \pm 2\sqrt{5}\) and another got \(\frac{10 \pm \sqrt{80}}{2}\), how could you check whether these are the same?”
\(\sqrt{80} = \sqrt{16 \cdot 5} = 4\sqrt{5}\), so \(\frac{10 \pm 4\sqrt{5}}{2} = 5 \pm 2\sqrt{5}\). They are the same.
🧑🏫 Spend 3–4 minutes here. Close with: "These skills — completing the square and the quadratic formula — are the foundation for everything that follows in this unit. Tomorrow we use them to solve equations whose solutions are complex numbers."
Cool-Down
5 min
Reference · How To
Oh, and Solve These Too
Reference · How To
Completing the Square
1. Half the coefficient of \(x\), then square it.
2. Add that value to both sides.
3. Factor as \(\left(x + \frac{b}{2}\right)^2 = k\).
4. Take \(\pm\) square root, isolate \(x\).
Quadratic Formula
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
1. Standard form \(ax^2 + bx + c = 0\).
2. Identify \(a, b, c\) — include signs.
3. Substitute and simplify.
Cool-Down
On your own — solve each equation
- \(x^2 + 12x = 13\)
- \(2x^2 - 5x + 1 = 0\)
🧑🏫 Problem 1 answer: \(x = 1\) or \(x = -13\)
Work: \(x^2 + 12x + 36 = 49\) → \((x + 6)^2 = 49\) → \(x + 6 = \pm 7\)
Problem 2 answer: \(x = \dfrac{5 \pm \sqrt{17}}{4}\)
Work: \(a = 2, b = -5, c = 1\). \(x = \frac{5 \pm \sqrt{25 - 8}}{4} = \frac{5 \pm \sqrt{17}}{4}\)
Common errors:
P1: Forgetting to add 36 to the right side too (gets \((x+6)^2 = 13\) instead of 49).
P2: Writing \(-b = -5\) instead of \(-(-5) = 5\).
Amplify · Activity Builder · Type-in
Setup (do once):
1. Add a Text component → include ALL student-facing text from the slide: "Solve each equation" and both problems.
2. Add two Math Response components → name them answer1 and answer2
3. Add a Note component below each → click </> → paste CL below
4. Straight quotes only — curly quotes break CL
Note for answer1:
content:
when answer1.submitted and answer1.latex = "1"
"That is one solution. What is the other? (x + 6) can also equal -7."
when answer1.submitted and answer1.latex = "-13"
"That is one solution. What is the other? (x + 6) can also equal 7."
when answer1.submitted and answer1.latex = "1,-13"
"Correct! x = 1 or x = -13."
when answer1.submitted and answer1.latex = "1, -13"
"Correct! x = 1 or x = -13."
when answer1.submitted and answer1.latex = "-13,1"
"Correct! x = 1 or x = -13."
when answer1.submitted and answer1.latex = "-13, 1"
"Correct! x = 1 or x = -13."
when answer1.submitted
"Complete the square: add 36 to both sides. What does (x + 6)^2 equal then?"
otherwise ""
Note for answer2:
content:
when answer2.submitted and answer2.latex = "\frac{5+\sqrt{17}}{4}"
"That is one solution. What is the other? Use the minus sign too."
when answer2.submitted and answer2.latex = "\frac{5-\sqrt{17}}{4}"
"That is one solution. What is the other? Use the plus sign too."
when answer2.submitted and answer2.latex = "\frac{5\pm\sqrt{17}}{4}"
"Correct! x = (5 +/- sqrt(17)) / 4."
when answer2.submitted and answer2.latex = "\frac{5 \pm \sqrt{17}}{4}"
"Correct! x = (5 +/- sqrt(17)) / 4."
when answer2.submitted and answer2.latex = "\frac{-5\pm\sqrt{17}}{4}"
"Almost! Check the sign of -b. Since b = -5, we get -(-5) = 5, not -5."
when answer2.submitted and answer2.latex = "\frac{-5 \pm \sqrt{17}}{4}"
"Almost! Check the sign of -b. Since b = -5, we get -(-5) = 5, not -5."
when answer2.submitted
"Use the quadratic formula with a = 2, b = -5, c = 1. What is b^2 - 4ac?"
otherwise ""