Lesson 19
Deriving the Quadratic Formula
19.1: Studying Structure (10 minutes)
Warmup
This warmup reminds students of the structure that governs the relationship between perfect squares written as squared factors and their equivalent expression in standard form. The key goal is for students to see that when we expand a squared factor of the form \((kx+m)^2\), the equivalent expression in standard form has this structure: \((kx)^2 + 2kmx+ m^2\). Seeing the coefficient of the squared term as \(\text {something}^2\) and the coefficient of the linear term as 2 times a product of two numbers (\(k\) and \(m\)) would enable students to complete the square more easily. This would in turn help them make sense of where the quadratic formula comes from, which they will explore in this lesson.
Students practice looking for and making use of structure as they think about the relationships between equivalent expressions (MP7).
Launch
Ask students to expand the squared factor \((x+4)^2\) into standard form and show all the steps. Students are likely to show \((x+4)(x+4) = x^2 + 4x + 4x + 4^2 = x^2 + 8x + 16\), or \((x+4)(x+4) = x^2 + 2(4x) + 4^2 = x^2 + 8x + 16\).
Next, arrange students in groups of 2. Tell students to use the relationship between the factored and standard forms to write some equivalent perfectsquare expressions. Ask students to think quietly about the expressions in the table before conferring with their partner.
Student Facing
Here are some perfect squares in factored and standard forms, and an expression showing how the two forms are related.

Study the first few examples, and then complete the missing numbers in the rest of the table.
factored form standard form \((x+4)^2\) \((1x)^2 + 2(\underline{\hspace{.2in}}x)(\underline{\hspace{.2in}}) + 4^2\) \(x^2 + 8x + 16\) \((2x + 5)^2\) \((2x)^2 + 2(\underline{\hspace{.2in}}x)(\underline{\hspace{.2in}}) + 5^2 \) \(4x^2 + 20x + 25 \) \((3x4)^2\) \((3x)^2 + 2(\underline{\hspace{0.2in}}x)(\underline{\hspace{0.2in}}) + (\underline{\hspace{0.2in}})^2 \) \(9x^2  24x + 16\) \((5x+\underline{\hspace{0.3in}})^2\) \((\underline{\hspace{.2in}}x)^2 + 2(\underline{\hspace{0.2in}}x)(\underline{\hspace{0.2in}}) + (\underline{\hspace{0.2in}})^2 \) \(25x^2 +30x + \underline{\hspace{0.3in}}\) \((kx+m)^2\) \((\underline{\hspace{.2in}}x)^2 + 2(\underline{\hspace{0.2in}}x)(\underline{\hspace{0.2in}}) + (\underline{\hspace{0.2in}})^2 \) \(\underline{\hspace{0.2in}}x^2 + \underline{\hspace{0.2in}}x + \underline{\hspace{0.2in}}\)  Look at the expression in the last row of the table. If \(ax^2 + bx+c\) is equivalent to \((kx+m)^2\), how are \(a, b\), and \(c\) related to \(k\) and \(m\)?
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Display the table for all to see and ask students for the missing values. Then, discuss the following questions to highlight the ways in which the numbers in the middle column are related to the two numbers in each expression in factored form.
 “For the expressions in the middle column, how do you know what to write for the first squared term?” (It’s the square of the linear term of the expression in factored form.)
 “The coefficient of the linear term is always 2 times the two other numbers. How do you know what those are?” (It’s the two terms from factored form: the linear term and the constant term.)
 “How do you know what to write for the last squared term?” (It’s the square of the constant term of the expression in factored form.)
 “When making \(25x^2 + 30x\) a perfect square, how do you know what values to write in the middle column?” (The first squared term is \((\text{something)}^2\) that equals \(25x^2\), so that “something” must be \(5x\) because \((5x)^2 = 25x^2\). The linear term is \(2(5x)(\text {another thing})\) that equals \(30x\), so that other thing must be 3 because \(2(5x)(3)=30x\). The last squared term is \(\text{(another thing)}^2\), so it is \(3^2\).)
Emphasize the general structure of the equivalent expressions, as shown in the last row. When \((kx+m)^2\) is expanded, the standard form always has the structure of \((kx)^2 + 2(kx)m + m^2\). Also, make sure students understand how the numbers in the factored form are related to \(ax^2+bx+c\), that is, that they recognize that \(a = k^2\), \(b = 2km\), and \(c=m^2\).
Highlight that when we want to complete the square for \(ax^2+bx\) and write an equivalent expression of the form \((kx+m)^2\), having a perfect square for \(a\) makes it much easier to find \(k\), and having an even number for \(b\) makes it easier to find \(m\) (because an even number gives a whole number when divided by 2).
19.2: Complete the Square using a Placeholder (10 minutes)
Activity
This is the first of two activities that help students derive the quadratic formula. Here students try to complete the square for a quadratic equation but without evaluating some of the numerical expressions along the way. Doing so enables them to recognize the parts of the quadratic formula but in numerical form (for example, instead of seeing \(\textb \pm \sqrt {b^24ac}\), they see \(\text5 \pm \sqrt{5^2  12}\)).
The linear coefficient of the given equation is an odd number. Students learn that there are strategies to transform the quadratic expression such that it has an even number for the linear coefficient and 1 for the leading coefficient, which makes it much easier to complete the square. These strategies involve multiplying the equation by a helpful factor and temporarily using simpler variables to stand in for complicated parts of an expression.
(The strategy of multiplying an equation by a number to make the \(a\) a perfect square is illustrated in an optional activity in the last lesson on completing the square. Students who completed that activity may recall this approach. Using a simple variable to stand in for a messier expression is also explored in an optional activity in an earlier lesson—the last lesson on rewriting expressions in factored form.)
Launch
The quadratic equation \(x^2 + 5x + 3 = 0\) is in the form of \(ax^2 + bx + c = 0\). What are the values of \(a\), \(b\), and \(c\)? Talk to a partner about whether it would be relatively simple to complete the square from this equation. (It’s doable, but probably pretty messy, as \(b\) is an odd number.)
Tell students that one way to make it easier to complete the square is to multiply the equation by a number such that the \(b\) is an even number and the \(a\) is a perfect square.
 Let’s try multiplying \(x^2 + 5x + 3 = 0\) by 2. We get \(2x^2 + 10x +6 = 0\). (The right side remains 0 because 0 times any number is 0.) We have 10, an even number for \(b\), but the \(a\) is not a perfect square.
 Let’s try multiplying it by 4. We get \(4x^2 + 20x + 12 = 0\). Now the \(a\) is a perfect square and the \(b\) is still an even number. To complete the square, it helps to isolate the existing constant and rewrite it as: \(4x^2 + 10x = \text12\).
 It’s still not immediately obvious what constant term to add to make a perfect square. It would be easier if the \(a\) or the coefficient of the squared term is 1. One way to deal with this is to think of \(4x^2\) as \(\text{(something)}^2\). What would the “something” be? (\(2x\))
 If we use a placeholder \(P\) to stand for \(2x\), we can write:
\(\displaystyle \begin {align} 4x^2 + 20x &= \text12\\ (2x)^2 + 10(2x) &= \text12\\P^2 + 10P &= \text12 \end{align}\)
Ask students to try completing the square for this simplerlooking equation.
Student Facing

One way to solve the quadratic equation \(x^2 + 5x + 3 = 0\) is by completing the square. A partially solved equation is shown here. Study the steps.
Then, knowing that \(P\) is a placeholder for \(2x\), continue to solve for \(x\) but without evaluating any part of the expression. Be prepared to explain each step.
\(\displaystyle \begin {align} x^2 + 5x + 3 &= 0 &\qquad& \text {Original equation}\\\\ 4x^2 + 20x + 12 &= 0 &\qquad& \text {Multiply each side by 4}\\\\ 4x^2 + 20x &= \text12 &\qquad& \text {Subtract 12 from each side}\\\\ (2x)^2 + 10(2x) &= \text12 &\qquad& \text {Rewrite }4x^2 \text{as }(2x)^2 \text { and }20x \text{ as }10(2x)\\\\ P^2 + 10P &= \text12 &\qquad& \text {Use } P \text{ as a placeholder for }2x\\\\ P^2 + 10P + \underline{\hspace{0.3in}}^2 &= \text12 + \underline{\hspace{0.3in}}^2\\\\ (P+\underline{\hspace{0.3in}})^2 &= \text12 + \underline{\hspace{0.3in}}^2\\\\ P+\underline{\hspace{0.3in}} &= \pm \sqrt {\text12 + \underline{\hspace{0.3in}}^2}\\\\ P &= \underline{\hspace{0.3in}} \pm \sqrt {\text12 + \underline{\hspace{0.3in}}^2}\\\\ P &= \underline{\hspace{0.3in}} \pm \sqrt {\underline{\hspace{0.3in}}^2  12}\\\\ 2x &= \underline{\hspace{0.3in}} \pm \sqrt {\underline{\hspace{0.3in}}^2  12}\\\\ x &= \\ \end {align}\)
 Explain how the solution is related to the quadratic formula.
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Activity Synthesis
Highlight the connections between the numbers in the solution \(x = \dfrac {\text5 \pm \sqrt {5^2  12}}{2}\) to the parameters in the original equation.
 The 5 in the expression is \(\textb\).
 The \(5^2\) is \(b^2\).
 The 12 is \(4(1)(3)\) or \(4ac\).
 The 2 in the denominator is \(2(1)\) or \(2a\).
Supports accessibility for: Visualspatial processing; Organization
19.3: Decoding the Quadratic Formula (15 minutes)
Activity
In the previous activity, students solved a quadratic equation by completing the square, but they did so without evaluating any of the numerical expressions. Students explained each step along the solving process and saw how it produced a solution that looks almost exactly like the quadratic formula, except that it contains numbers instead of \(a, b,\) and \(c\). That activity gave students concrete insights into the process of deriving the quadratic formula, preparing them to do the same in more abstract terms.
In this activity, students study a series of steps taken to solve \(ax^2+bx+c=0\) by completing the square and make sense of how it leads to the quadratic formula. Along the way, they see that the solving process involves similar maneuvers as those seen earlier, for example, multiplying the equation by a factor to make the leading coefficient a perfect square, writing the linear coefficient as 2 times two numbers, and adding the square of one of the numbers to complete the square. They see that the result of solving the equation by completing the square is the quadratic formula.
Launch
Tell students that they will now study a workedout solution to \(ax^2+bx+c=0\). There are no numbers in this equation, but the process of solving should be familiar. Ask students to analyze the solution and record an explanation for each step. (They should explain why each step is taken and not only what happens in each step.)
Design Principle(s): Support sensemaking; Cultivate conversation
Supports accessibility for: Organization; Attention
Student Facing
Here is one way to make sense of how the quadratic formula came about. Study the derivation until you can explain what happened in each step. Record your explanation next to each step.
\(\displaystyle \begin {align} ax^2 + bx + c &= 0\\\\ 4a^2x^2 + 4abx + 4ac &= 0\\\\ 4a^2x^2+4abx &=\text4ac\\\\ (2ax)^2 + 2b(2ax) &= \text4ac\\\\ M^2 + 2bM &= \text4ac\\\\ M^2 + 2bM + b^2 &= \text4ac + b^2\\\\ (M+b)^2 &= b^24ac\\\\ M+b &= \pm \sqrt{b^24ac}\\\\ M &= \textb \pm \sqrt{b^24ac}\\\\ 2ax &= \textb \pm \sqrt{b^24ac}\\\\ x &= \frac {\textb \pm \sqrt{b^24ac}}{2a} \end {align}\)
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Student Facing
Are you ready for more?
Here is another way to derive the quadratic formula by completing the square.
 First, divide each side of the equation \(ax^2+bx+c=0\) by \(a\) to get \(x^2+\frac {b}{a}x+\frac{c}{a}=0\).
 Then, complete the square for \(x^2+\frac {b}{a}x+\frac{c}{a}=0\).

The beginning steps of this approach are shown here. Briefly explain what happens in each step.
\(\begin{align} x^2 + \frac{b}{a}x + \frac{c}{a} &= 0 &\qquad& \text{Original equation}\\ x^2 + \frac{b}{a}x &= \text\frac{c}{a} &\qquad& \text[1]\\ x^2 + 2\left(\frac {b}{2a}\right)x + \left(\frac {b}{2a}\right)^2&=\text\frac ca + \left(\frac {b}{2a}\right)^2 &\qquad& \text[2]\\ \left(x+\frac {b}{2a}\right)^2&= \text \frac ca + \frac {b^2}{4a^2} &\qquad& \text[3]\\ \left(x+\frac {b}{2a}\right)^2&= \text \frac {4ac}{4a^2} + \frac {b^2}{4a^2} &\qquad& \text[4]\\ \left(x+\frac {b}{2a}\right)^2&= \frac {b^24ac}{4a^2} &\qquad& \text[5]\\ x+\frac{b}{2a} &= \pm \sqrt { \frac {b^24ac}{4a^2}} &\qquad& \text[6]\\ x+\frac{b}{2a} &= \pm \frac {\sqrt {b^24ac}}{\sqrt{4a^2}} &\qquad& \text[7] \end{align}\)
 Continue the solving process until you have the equation \(x = \dfrac{\textb \pm \sqrt{b^24ac}}{2a}\).
Student Response
Student responses to this activity are available at one of our IM Certified Partners
Anticipated Misconceptions
Some students may think that we multiply just by 4 and wonder where the \(a\) comes from. Point out that we need the first term to be a perfect square. In the previous activity, the \(a\) value was 1 (which was already a perfect square). We can’t be certain that \(a\) is a perfect square, so we multiply by 4 and \(a\) to make the first term \(4a^2x^2\), which is \((2ax)^2\), a perfect square.
Some may wonder where the \(M\) comes from. Consider displaying a completed solution from the previous activity that parallels this work. Remind students of the placeholder \(P\) used there. Invite them to compare other steps in the previous activity to those in this activity to help them explain the derivation.
Activity Synthesis
Invite students to share their explanations for each step. Highlight a few key maneuvers:

Multiplying the equation by \(4a\) makes the coefficient of \(x^2\) a perfect square and the coefficient of the linear term an even number, both of which make completing the square much easier.
Some students may wonder why \(4a\) is chosen to complete the square when solving \(ax^2+bx+c=0\). Why not \(a\), \(9a\), or \(16a\)? Any of these would work. The equation could also have been divided by \(a\) to make the first term \(x^2\), which is a perfect square. (If time permits, consider asking students to try completing the square using one of these alternatives.) While these alternatives result in a perfect square for the first term, they may not give an even number for the coefficient of the second term or they may produce an equation with larger numbers, making it a bit trickier to manipulate.
 We can tell what constant term is needed to complete the square by dividing the coefficient of the linear term by the positive square root of the coefficient of \(x^2\), taking half of it, and squaring that number. In this case:
 The coefficient of \(x\) is \(4ab\).
 The coefficient of \(x^2\) is \(4a^2\), so the square root is \(2a\).
 Dividing \(4ab\) by \(2a\) gives \(2b\). Half of \(2b\) is \(b\).
 The constant term that completes the square is \(b^2\).
Emphasize that the quadratic formula essentially captures the steps for completing the square in one expression. Every time we solve a quadratic equation by completing the square, we are essentially using the quadratic formula, but in a less condensed way.
Lesson Synthesis
Lesson Synthesis
Students are not expected to internalize how to derive the quadratic formula by the end of the lesson. They should, however, understand that each part of the complicated expression can be traced back to a step in completing the square. In other words, they should recognize that the formula is not an isolated, opaque method that mysteriously produces solutions to quadratic equations.
Discuss with students:
 “How has your understanding of the quadratic formula changed from the work in this lesson?”
 “A classmate who is absent today is not sure where the quadratic formula came from. What would you say to help them understand, in a nutshell, what the formula is all about?”
 “If the formula is connected to the steps of completing the square, why not just complete the square when we need to solve equations? Why do you think is it helpful to have a formula, even if it involves quite a few operations?”
19.4: Cooldown  Step by Step (5 minutes)
CoolDown
Cooldowns for this lesson are available at one of our IM Certified Partners
Student Lesson Summary
Student Facing
Recall that any quadratic equation can be solved by completing the square. The quadratic formula is essentially what we get when we put all the steps taken to complete the square for \(ax^2 + bx+c =0\) into a single expression.
When we expand a squared factor like \((3x+5)^2\), the result is \((3x)^2 + 2(5)(3x) + 25\). Notice how the \((3x)\) appears in two places. If we replace \((3x)\) with another letter like \(P\), we have \(P^2+10P+25\), which is a recognizable perfect square.
Likewise, if we expand \((kx+m)^2\), we have \((kx)^2+2m(kx)+m^2\). Replacing \(kx\) with \(P\) gives \(P^2+2mP+m^2\), also a recognizable perfect square.
To complete the square is essentially to make one side of the equation have the same structure as \((kx)^2 + 2m(kx) + m^2\). Substituting a letter for \((kx)\) makes it easier to see what is needed to complete the square. Let’s complete the square for \(ax^2 + bx+c =0\)!
 Start by subtracting \(c\) from each side.
\(\displaystyle ax^2 + bx = \textc\)
 Next, let’s multiply both sides by \(4a\). On the left, this gives \(4a^2\), a perfect square for the coefficient of \(x^2\).
\(\displaystyle 4a^2x^2 +4abx = \text4ac\)
 \(4a^2x^2\) can be written \((2ax)^2\), and \(4abx\) can be written \(2b(2ax)\).
\(\displaystyle (2ax)^2 + 2b(2ax) = \text4ac\)
 Let’s replace \((2ax)\) with the letter \(P\).
\(\displaystyle P^2 + 2bP= \text4ac\)
 \(b^2\) is the constant term that completes the square, so let’s add \(b^2\) to each side.
\(\displaystyle P^2 + 2bP + b^2 = \text4ac\ + b^2\)
 The left side is now a perfect square and can be written as a squared factor.
\(\displaystyle (P + b)^2 = b^2 4ac\)

The square roots of the expression on the right are the values of \(P+b\).
Once \(P\) is isolated, we can write \(2ax\) in its place and solve for \(x\).
\(\displaystyle \begin {align} P+b &= \pm \sqrt {b^24ac} \\ \\ P &= \textb \pm \sqrt {b^24ac}\\ \\2ax &= \textb \pm \sqrt {b^24ac}\end{align}\)

The solution is the quadratic formula!
\(\displaystyle \begin {align}\quad x &= \dfrac{\textb \pm \sqrt {b^24ac}}{2a}\end{align}\)