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.
Warm-Up
5 min
Find the Perfect Squares
The expression \(x^2 + 8x + 16\) is equivalent to \((x + 4)^2\). Are these expressions equivalent to \((x + n)^2\) for some number \(n\)? If so, identify the number, \(n\).
On your own
  1. \(x^2 + 10x + 25\)
  2. \(x^2 + 10x + 100\)
  3. \(x^2 + 7x + 12\)
  4. \(x^2 - 6x + 9\)
🧑‍🏫 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.
Warm-Up Synthesis
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?”
Sample Responses click to reveal ▶
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
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: \(x^2 - 6x + 7 = 0\)
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.
Activity 1 · Work Time
Different Ways to Solve It: \(x^2 - 6x + 7 = 0\)
Elena — Completing the Square
“First I added 2 to each side:”
\(x^2 - 6x + 7 + 2 = 0 + 2\)
\(x^2 - 6x + 9 = 2\)
“So that tells me:”
\((x - 3)^2 = 2\)
“I can find the square roots of both sides:”
\(x - 3 = \pm\sqrt{(2)}\)
“Which is the same as:”
\(x = 3 \pm \sqrt{(2)}\)
Han — Quadratic Formula
“I used the quadratic formula:”
\(x = \dfrac{-b \pm \sqrt{(b^2 - 4ac)}}{2a}\)
“Since \(a = 1\), \(b = -6\), \(c = 7\):”
\(x = \dfrac{-(-6) \pm \sqrt{((-6)^2 - 4(1)(7))}}{2(1)}\)
“Which gives me:”
\(x = \dfrac{6 \pm \sqrt{(36 - 28)}}{2}\)
\(x = \dfrac{6 \pm \sqrt{(8)}}{2}\)
Discuss with your partner
  1. Do you agree with either of them? Explain your reasoning.
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.
Activity 1 · Synthesis
Do both methods give the same solutions?
“The solutions look different. How can you tell if they are the same values?”
Elena — Completing the Square
Add 2 so the constant becomes \((-3)^2 = 9\):
\(x^2 - 6x + 7 + 2 = 0 + 2\)
Her answer
\(x = 3 \pm \sqrt{(2)}\)
Han — Quadratic Formula
\(a=1,\; b=-6,\; c=7\):
\(x = \dfrac{-(-6) \pm \sqrt{((-6)^2 - 4(1)(7))}}{2(1)}\)
His answer
\(x = \dfrac{6 \pm \sqrt{(8)}}{2}\)
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
Completing the Square
Math · We Say · Meaning
★ 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.
Reference · How To
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 \(x\) — add 3 to both sides \(x - 3 + 3 = \pm\sqrt{(2)} + 3\)
\(x = 3 \pm \sqrt{(2)}\)
🧑‍🏫 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.
MASL · Language & Notation
The Plus-or-Minus Symbol
Math · We Say · Meaning
★ 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
The Quadratic Formula
Math · We Say · Meaning
★ 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.
Reference · How To
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 \(x = \dfrac{-1 + 7}{6} = \dfrac{6}{6} = 1\)  or  \(x = \dfrac{-1 - 7}{6} = \dfrac{-8}{6} = -\dfrac{4}{3}\)
★ 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.
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.
Amplify · Activity Builder · Type-in

Setup (do once):

1. Add a Text component → include ALL student-facing text: slide heading and all 6 problems.

2. Add six Math Response components → name them answer1 through answer6.

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

4. Straight quotes only — curly quotes break CL.

Problem 1: (x − 5)² = 36
content: when answer1.submitted and answer1.latex = "11,-1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "11, -1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "-1,11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "-1, 11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "11;-1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "11; -1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "-1;11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "-1; 11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=11,x=-1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=11, x=-1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=-1,x=11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=-1, x=11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=11;x=-1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=11; x=-1" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=-1;x=11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=-1; x=11" "Correct! x = 11 or x = -1." when answer1.submitted and answer1.latex = "x=11" "That is one solution. What is the other? (x - 5) can also equal -6." when answer1.submitted and answer1.latex = "x=-1" "That is one solution. What is the other? (x - 5) can also equal 6." when answer1.submitted and answer1.latex = "11" "That is one solution. What is the other? (x - 5) can also equal -6." when answer1.submitted and answer1.latex = "-1" "That is one solution. What is the other? (x - 5) can also equal 6." when answer1.submitted "Take the square root of both sides: x - 5 = ±6. What are the two values of x? (Enter both: x = ?, x = ?)" otherwise ""
Problem 2: (x + 3)² = 10
content: when answer2.submitted and answer2.latex = "-3\pm\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3 \pm \sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3+\sqrt{10},-3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3+\sqrt{10}, -3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3-\sqrt{10},-3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3-\sqrt{10}, -3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3+\sqrt{10};-3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3+\sqrt{10}; -3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3-\sqrt{10};-3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "-3-\sqrt{10}; -3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3+\sqrt{10},x=-3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3+\sqrt{10}, x=-3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3-\sqrt{10},x=-3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3-\sqrt{10}, x=-3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3+\sqrt{10};x=-3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3+\sqrt{10}; x=-3-\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3-\sqrt{10};x=-3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3-\sqrt{10}; x=-3+\sqrt{10}" "Correct! x = -3 ± sqrt(10)." when answer2.submitted and answer2.latex = "x=-3+\sqrt{10}" "That is one solution. What is the other? Use the minus sign too." when answer2.submitted and answer2.latex = "x=-3-\sqrt{10}" "That is one solution. What is the other? Use the plus sign too." when answer2.submitted and answer2.latex = "-3+\sqrt{10}" "That is one solution. What is the other? Use the minus sign too." when answer2.submitted and answer2.latex = "-3-\sqrt{10}" "That is one solution. What is the other? Use the plus sign too." when answer2.submitted and answer2.latex = "3\pm\sqrt{10}" "Almost! Check the sign: x + 3 = ±sqrt(10), so x = -3 ± sqrt(10), not +3." when answer2.submitted and answer2.latex = "3 \pm \sqrt{10}" "Almost! Check the sign: x + 3 = ±sqrt(10), so x = -3 ± sqrt(10), not +3." when answer2.submitted "Take the square root of both sides: x + 3 = ±sqrt(10). Now isolate x. (Give your answer as x = -3 ± sqrt(10), or enter both values: x = ?, x = ?)" otherwise ""
Problem 3: x² + 8x + 16 = 0
content: when answer3.submitted and answer3.latex = "-4" "Correct! This is a perfect square with one solution: x = -4." when answer3.submitted and answer3.latex = "-4,-4" "Good start — but a double root counts as one solution. Write just x = -4." when answer3.submitted and answer3.latex = "-4, -4" "Good start — but a double root counts as one solution. Write just x = -4." when answer3.submitted and answer3.latex = "4" "Almost! Factor the left side: (x + 4)^2 = 0. What value of x makes x + 4 = 0?" when answer3.submitted "Factor the left side — it is a perfect square trinomial. What does it factor to?" otherwise ""
Problem 4: x² + 14x + 40 = 0
content: when answer4.submitted and answer4.latex = "-4,-10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-4, -10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-10,-4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-10, -4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-4;-10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-4; -10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-10;-4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "-10; -4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-4,x=-10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-4, x=-10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-10,x=-4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-10, x=-4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-4;x=-10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-4; x=-10" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-10;x=-4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-10; x=-4" "Correct! x = -4 or x = -10." when answer4.submitted and answer4.latex = "x=-4" "That is one solution. What is the other? The equation has two." when answer4.submitted and answer4.latex = "x=-10" "That is one solution. What is the other? The equation has two." when answer4.submitted and answer4.latex = "-4" "That is one solution. What is the other? The equation has two." when answer4.submitted and answer4.latex = "-10" "That is one solution. What is the other? The equation has two." when answer4.submitted "Complete the square: half of 14 is 7, and 7^2 = 49. Add 49 to both sides. What does (x + 7)^2 equal? (Enter both: x = ?, x = ?)" otherwise ""
Problem 5: x² + 3x − 5 = 0
content: when answer5.submitted and answer5.latex = "\frac{-3\pm\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3 \pm \sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3+\sqrt{29}}{2},\frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3+\sqrt{29}}{2}, \frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3-\sqrt{29}}{2},\frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3-\sqrt{29}}{2}, \frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3+\sqrt{29}}{2};\frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3+\sqrt{29}}{2}; \frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3-\sqrt{29}}{2};\frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "\frac{-3-\sqrt{29}}{2}; \frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3+\sqrt{29}}{2},x=\frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3+\sqrt{29}}{2}, x=\frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3-\sqrt{29}}{2},x=\frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3-\sqrt{29}}{2}, x=\frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3+\sqrt{29}}{2};x=\frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3+\sqrt{29}}{2}; x=\frac{-3-\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3-\sqrt{29}}{2};x=\frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3-\sqrt{29}}{2}; x=\frac{-3+\sqrt{29}}{2}" "Correct! x = (-3 ± sqrt(29)) / 2." when answer5.submitted and answer5.latex = "x=\frac{-3+\sqrt{29}}{2}" "That is one solution. What is the other? Include the minus version too." when answer5.submitted and answer5.latex = "x=\frac{-3-\sqrt{29}}{2}" "That is one solution. What is the other? Include the plus version too." when answer5.submitted and answer5.latex = "\frac{-3+\sqrt{29}}{2}" "That is one solution. What is the other? Include the minus version too." when answer5.submitted and answer5.latex = "\frac{-3-\sqrt{29}}{2}" "That is one solution. What is the other? Include the plus version too." when answer5.submitted and answer5.latex = "\frac{3\pm\sqrt{29}}{2}" "Almost! Check -b: since b = 3, we get -b = -3, not 3." when answer5.submitted and answer5.latex = "\frac{3 \pm \sqrt{29}}{2}" "Almost! Check -b: since b = 3, we get -b = -3, not 3." when answer5.submitted "Use the quadratic formula with a = 1, b = 3, c = -5. What is b^2 - 4ac? (Give your answer as x = (-3 ± sqrt(29)) / 2, or enter both values: x = ?, x = ?)" otherwise ""
Problem 6: 3x² + x − 4 = 0
content: when answer6.submitted and answer6.latex = "1,-\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "1, -\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "-\frac{4}{3},1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "-\frac{4}{3}, 1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "1;-\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "1; -\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "-\frac{4}{3};1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "-\frac{4}{3}; 1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=1,x=-\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=1, x=-\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=-\frac{4}{3},x=1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=-\frac{4}{3}, x=1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=1;x=-\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=1; x=-\frac{4}{3}" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=-\frac{4}{3};x=1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=-\frac{4}{3}; x=1" "Correct! x = 1 or x = -4/3." when answer6.submitted and answer6.latex = "x=1" "That is one solution. What is the other? The discriminant is a perfect square — check your work." when answer6.submitted and answer6.latex = "x=-\frac{4}{3}" "That is one solution. What is the other? The discriminant is a perfect square — check your work." when answer6.submitted and answer6.latex = "1" "That is one solution. What is the other? The discriminant is a perfect square — check your work." when answer6.submitted and answer6.latex = "-\frac{4}{3}" "That is one solution. What is the other? The discriminant is a perfect square — check your work." when answer6.submitted "Use the quadratic formula with a = 3, b = 1, c = -4. The discriminant is 1 + 48 = 49. What is sqrt(49)? (Enter both: x = ?, x = ?)" otherwise ""
Lesson Synthesis
Different answers, same solution?
Different methods can give answers that look different but are equivalent. Simplify radicals and fractions to verify.
“If one person got \(5 \pm 2\sqrt{(5)}\) and another got \(\dfrac{10 \pm \sqrt{(80)}}{2}\), how could you check whether these are the same?”
Sample Response click to reveal ▶
\(\sqrt{(80)} = \sqrt{(16 \cdot 5)} = 4\sqrt{(5)}\), so \(\dfrac{10 \pm 4\sqrt{(5)}}{2} = 5 \pm 2\sqrt{(5)}\). They are the same.
🧑‍🏫 Use this to bridge comparing answers from Activity 2. If two pairs got different-looking answers for the same problem, show them here how to verify equivalence.
Activity 2 · Synthesis
Which method works best when?
For problems where \(ax^2 + bx + c = 0\)
Square roots
When: no \(x\) term — equation is (or can be written as) \((\text{expr})^2 = k\)
\((x - 5)^2 = 36\)
Completing the square
When: \(a = 1\) and b is even
\(x^2 - 6x + 7 = 0\)
Check:
\(a = 1\) ✓
\(b = -6\) and half of −6 is −3 ✓
Quadratic formula
When: \(a \neq 1\), b is odd, or coefficients are messy — it always works
\(2x^2 - 5x + 1 = 0\)
Why is this "messy"?
Divide by 2 first: \(x^2 - \tfrac{5}{2}x + \tfrac{1}{2} = 0\)
Half of \(-\tfrac{5}{2}\) is \(-\tfrac{5}{4}\), square is \(\tfrac{25}{16}\) — fractions pile up.
QF gives \(x = \dfrac{5 \pm \sqrt{(17)}}{4}\) directly.
🧑‍🏫 Answers: 1. Completing the square (a=1, b even: half of 4 is 2). 2. Quadratic formula (a=3, messy to complete the square). 3. Square roots (already in (expr)²=k form). 4. Completing the square (a=1, b even: half of −8 is −4). Select 2–3 pairs to share. Ask: "What was the first thing you looked at to decide?"
Lesson Synthesis
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 \(ax^2 + bx + c = 0\).
Try Saying “I would use ___ when ___.”
Try Saying “I can check whether they are the same by ___.”
🧑‍🏫 Spend 2–3 minutes here. Then continue to the equivalence check on the next slide.
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
  1. \(x^2 + 12x = 13\)
  2. \(2x^2 - 5x + 1 = 0\)
🧑‍🏫 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 ""