Lesson 16
The Quadratic Formula
16.1: Evaluate It (5 minutes)
Warmup
This warmup prompts students to evaluate the kinds of numerical expressions they will see in the lesson. The expressions involve rational square roots, fractions, and the \(\pm\) notation.
As students work, notice any common errors or challenges so they can be addressed during the class discussion.
Launch
Tell students to evaluate the expressions without using a calculator.
Student Facing
Each expression represents two numbers. Evaluate the expressions and find the two numbers.
 \(1 \pm \sqrt{49}\)
 \(\displaystyle \frac{8 \pm 2}{5}\)
 \(\pm \sqrt{(\text5)^24 \boldcdot 4 \boldcdot 1}\)
 \(\displaystyle \frac{\text18 \pm \sqrt{36}}{2 \boldcdot 3}\)
Student Response
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Anticipated Misconceptions
Students may be unfamiliar with evaluating rational expressions in which the numerator contains more than one term. To help students see the structure of the expressions, consider decomposing them into a sum of two fractions. For example, show that \(\frac{8\pm2}{5}\) can be written as \(\frac85\pm\frac25\). This approach can also help to avoid a common error of dividing only the first term by the denominator (\(\frac{4+7}{2}\ne2+7\)). Some students may incorrectly write \(\frac{\sqrt{36}}{2}\) as \(\sqrt{18}\). Point out that the first expression is equal to 3 while the other has to be greater than 3 since \(\sqrt{18}\approx4.243\).
Activity Synthesis
Select students to share their responses and reasoning. Address any common errors. As needed, remind students of the properties and order of operations and the meaning and use of the \(\pm\) symbol.
16.2: Pesky Equations (10 minutes)
Activity
In this activity, students encounter equations that are challenging to solve using the methods they have learned, motivating students to seek a more efficient method.
Launch
Arrange students in groups of 2. Ask partners to choose the same equation. Give students quiet time to solve the equation and then time to discuss their solutions and strategy. If they finish solving their chosen equation, ask them to choose another one to solve. Leave a few minutes for a wholeclass discussion.
Provide access to calculators for numerical computations.
Design Principle(s): Support sensemaking
Supports accessibility for: Socialemotional skills; Conceptual processing
Student Facing
Choose one equation to solve, either by rewriting it in factored form or by completing the square. Be prepared to explain your choice of method.
 \(x^22x1.25=0\)
 \(5x^2+9x44=0\)
 \(x^2+1.25x=0.375\)
 \(4x^228x+29=0\)
Student Response
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Activity Synthesis
Consider arranging students who solved the same equation in groups of 2 to 3 to discuss their strategies and then displaying the correct solutions for all to see.
Invite students to share their reflections on the solving process. Discuss questions such as:
 “What method did you choose and why?”
 “Did you find yourself choosing one method and then switching to another? If so, what prompted you to do that?”
 “Did you run into any challenges when rewriting the equation (or completing the square)? What were some of the challenges?”
Acknowledge that all of these equations are cumbersome to solve by either rewriting in factored form or completing the square. The last equation cannot be written in factored form (with rational coefficients), so completing the square is the only way to go. Tell students they are about to learn a formula that gives the solutions to any quadratic equation.
16.3: Meet the Quadratic Formula (20 minutes)
Activity
This activity introduces the quadratic formula. Students begin by applying the formula and using it to solve various equations, ranging from those that can be easily solved using other methods to the kinds that would be quite tedious to solve without the formula. They then verify that the formula gives the same solutions as those calculated by another method. Students notice that, for equations that cannot be easily rewritten using factored form or solved by completing the square, the formula offers quite an efficient way to find the solutions.
Launch
Display the equation \(ax^2 + bx + c =0\) for all to see. Tell students that \(a\), \(b\), and \(c\) are numbers and \(a\) is not 0. Ask students if they could complete the square for this equation without first replacing \(a\), \(b\), and \(c\) with numbers.
Explain that this can indeed be done! The outcome of completing the square is not going to be numerical solutions (because no numbers are used), but rather a general formula for finding the solutions of the quadratic equation. While students won’t have to complete the square for this equation now, they will see the formula and try using it.
Display the quadratic formula for all to see. Tell students that when an equation is of the form \(ax^2 + bx + c =0\), where \(a\), \(b\), and \(c\) are numbers and \(a\) is not 0, we can find its solutions by using the formula: \(\displaystyle x=\dfrac{\text b \pm \sqrt{b^24ac}}{2a}\)
Guide students through the steps of using the formula to find the solutions to \(x^2  8x + 15=0\).
 “First, what do we expect the solutions to be?” (3 and 5, because we can rewrite it as \((x3)(x5)=0\).)
 “If \(x^2  8x + 15\) is \(ax^2 + bx + c\), what are the values of \(a\), \(b\), and \(c\)?” (\(a=1,b=\text8,c=15\).)
 Write out the formula as is.

Replace \(a\), \(b\), and \(c\) with the corresponding numbers from the equation:
\(\displaystyle x=\dfrac{\text (\text8) \pm \sqrt{(\text8)^24(1)(15)}}{2(1)}\)
 Evaluate one part of the expression at a time, ending with \(x = 3\) and \(x=5\).
Provide access to calculators for numerical computations.
If time is limited, ask students to complete at least 2 equations, including an equation in which the leading coefficient is not 1.
Supports accessibility for: Organization; Attention; Socialemotional skills
Student Facing
Here is a formula called the quadratic formula.
\(\displaystyle x=\dfrac{\text b \pm \sqrt{b^24ac}}{2a}\)
The formula can be used to find the solutions to any quadratic equation in the form of \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are numbers and \(a\) is not 0.
This example shows how it is used to solve \(x^2  8x + 15=0\), in which \(a=1\), \(b=\text8\), and \(c=15\).
\(\displaystyle \begin {align} x &=\dfrac{\text b \pm \sqrt{b^24ac}}{2a} &\qquad &\text{original equation}\\ x &=\dfrac{\text (\text8) \pm \sqrt{(\text8)^24(1)(15)}}{2(1)} &\qquad &\text{substitute the values of }a, b, \text{and }c &\\ x &=\dfrac{8 \pm \sqrt{6460}}{2} &\qquad &\text {evaluate each part of the expression} \\ x &=\dfrac{8 \pm \sqrt{4}}{2} \\x &=\dfrac{8 \pm 2}{2}\\ x &=\frac{10}{2} \qquad \text {or} \qquad x =\frac{6}{2} \\x &=\text{ }5 \qquad \text{ } \text {or} \qquad x =\text{ }3 \end{align}\)
Here are some quadratic equations and their solutions. Use the quadratic formula to show that the solutions are correct.
 \(x^2 + 4x  5 = 0\). The solutions are \(x=\text5\) and \(x=1\).
 \(x^2 + 7x + 12 = 0\). The solutions are \(x=\text3\) and \(x=\text4\).
 \(x^2 + 10x + 18 = 0\). The solutions are \(x={\text5} \pm \frac{\sqrt{28}}{2}\).
 \(x^2  8x + 11 = 0\). The solutions are \(x=4 \pm\frac{ \sqrt{20}}{2}\).
 \(9x^26x+1=0\). The solution is \(x=\frac13\).
 \(6x^2+9x15=0\). The solutions are \(x=\text\frac52\) and \(x=1\).
Student Response
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Student Facing
Are you ready for more?
 Use the quadratic formula to solve \(ax^2+c=0\). Let’s call the resulting equation P.

Solve the equation \(3x^227=0\) in two ways, showing your reasoning for each:
 Without using any formulas.
 Using equation P.

Check that you got the same solutions using each method.
 Use the quadratic formula to solve \(ax^2+bx=0\). Let’s call the resulting equation Q.

Solve the equation \(2x^2+5x=0\) in two ways, showing your reasoning for each:
 Without using any formulas.
 Using equation Q.
 Check that you got the same solutions using each method.
Student Response
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Activity Synthesis
Much of the student discussion will have happened in small groups. Focus the wholeclass conversation on whether the quadratic formula works for solving all equations and when it might be a preferred method. Ask students,
 “Look at the list of equations and the work you did to solve them. Do certain equations lend themselves to certain methods of solving? Why?” (We can see the factors of some expressions right away, so rewriting it in factored form would be the quickest method to solve the equations. Other expressions cannot be easily rewritten in factored form, but can be easily transformed into perfect squares. For example, when the coefficient of \(x^2\) is 1, the coefficient of the linear term is an even number, and the constant term is also a whole number, completing the square is straightforward. When the squared term has a coefficient other than 1, the other two methods are less practical, so the quadratic formula may be the quickest approach.)
 “Why do you think the quadratic formula is useful only when the \(a\) in \(ax^2+bx+c=0\) is not 0?” (If \(a\) is 0, the expression is no longer quadratic because the squared term disappears, leaving only \(bx+c\), which is linear.)
Select students who used the quadratic formula to solve the last few equations to explain their solutions and display their work for all to see. Discuss any challenges or disagreements in using the formula.
Tell students that they will use the formula to solve other equations and find out more about its merits and how it compares to other methods of solving.
Design Principle(s): Support sensemaking; Maximize metaawareness
Lesson Synthesis
Lesson Synthesis
Display the equation \(25x^250x+16=0\) for all to see. Ask students how they prefer to solve it (by rewriting the expression in factored form, completing the square, or using the quadratic formula) and why. Then, ask them to solve the equation using their preferred method.
Possible explanations for the different methods:

Rewriting in factored form: The equation can be rewritten as \((5x8)(5x2)=0\) and solved using the zero product property.

Completing the square: \(25x^250x+25\) is a perfect square, so we can just add 9 to either side of the original equation and find the square root of 9.

The quadratic formula, because it always works.
The solutions (by any method) are \(\frac25\) and \(\frac85\).
16.4: Cooldown  Solving and Checking (5 minutes)
CoolDown
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Student Lesson Summary
Student Facing
We have learned a couple of methods for solving quadratic equations algebraically:
 by rewriting the equation as \(\text{factored form}=0\) and using the zero product property
 by completing the square
Some equations can be solved quickly with one of these methods, but many cannot. Here is an example: \(5x^23x1=0\). The expression on the left cannot be rewritten in factored form with rational coefficients. Because the coefficient of the squared term is not a perfect square, and the coefficient of the linear term is an odd number, completing the square would be inconvenient and would result in a perfect square with fractions.
The quadratic formula can be used to find the solutions to any quadratic equation, including those that are tricky to solve with other methods.
For an equation of the form \(ax^2+bx+c=0\), where \(a\), \(b\), and \(c\) are numbers and \(a \neq 0\), the solutions are given by:
\(\displaystyle x=\dfrac{\text b \pm \sqrt{b^24ac}}{2a}\)
For the equation \(5x^23x1=0\), we see that \(a=5\), \(b=\text3\), and \(c=\text1\). Let’s solve it!
\(\displaystyle \begin {align} x &=\dfrac{\text b \pm \sqrt{b^24ac}}{2a} &\qquad &\text{the quadratic formula}\\ x &=\dfrac{\text(\text3) \pm \sqrt{(\text3)^24(5)(\text1)}}{2(5)} &\qquad &\text{substitute the values of }a, b, \text{and }c &\\ x &=\dfrac{3 \pm \sqrt{9+20}}{10} &\qquad &\text {evaluate each part of the expression} \\ x &=\dfrac{3 \pm \sqrt{29}}{10}\end{align}\)
A calculator gives approximate solutions of 0.84 and 0.24 for \(\frac{3 + \sqrt{29}}{10}\) and \(\frac{3  \sqrt{29}}{10}\).
We can also use the formula for simpler equations like \(x^29x+8=0\), but it may not be the most efficient way. If the quadratic expression can be easily rewritten in factored form or made into a perfect square, those methods may be preferable. For example, rewriting \(x^29x+8=0\) as \((x1)(x8)=0\) immediately tells us that the solutions are 1 and 8.