Lesson 15
Quadratic Equations with Irrational Solutions
15.1: Roots of Squares (5 minutes)
Warm-up
This warm-up reminds students that we can use the notation \(\sqrt{A}\) to express the side length of a square with area \(A\) and that the value of \(\sqrt{A}\) may be not be a whole number or a fraction.
To find the area of a tilted square, students might choose different strategies:
- Decompose the square and rearrange the parts into rectangles.
- Enclose the tilted square with another square that is on the grid and subtract the areas of the extra triangles.
- Use the Pythagorean Theorem to calculate the squared value of the side length.
Launch
Display the entire task for all to see. Give students 3 minutes of quiet think time. Select students to share their responses and how they reasoned about the side length and area of each square.
Student Facing
Here are some squares whose vertices are on a grid.
Find the area and the side length of each square.
| square | area (square units) | side length (units) |
|---|---|---|
| A | ||
| B | ||
| C |
Student Response
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Activity Synthesis
Invite students to share their solutions. It is not essential to discuss how students found the area. What is important is that students recall that we can use the square root notation \(\sqrt 9\) to refer to the side length of a square with area 9 square units and that \(\sqrt 9=3\). When the area is 2 or 10 square units, the square root of each number is not a whole number or a fraction, but we can write \(\sqrt 2\) and \(\sqrt{10}\) rather than writing their decimal approximations.
Remind students that any positive number has two square roots because there are two numbers (one positive and one negative) that, when squared, give that number. Also remind students that the \(\sqrt{\phantom{3}}\) symbol refers only to the positive square root of a number. In this context, if \(s\) is the side length of a square and \(A\) its area, we can describe the relationship between the two quantities as \(s = \sqrt A\) because only positive side lengths make sense.
15.2: Solutions Written as Square Roots (15 minutes)
Activity
This activity introduces the use of \(\pm\) notation as a simple way to express the two square roots of a number. Students solve several simple equations by finding square roots and express their solutions using the notation.
Students also see that sometimes the solutions are not rational numbers and can be expressed exactly using the radical notation rather than using their decimal approximations. For example, the solutions to \(x^2=24\) can be written as \(x= \pm\sqrt{24}\) rather than \(x\approx \pm 4.9\) and the solutions to \((x-3)^2 =24\) can be written as \(x=\pm \sqrt {24} +3\) instead of \(x\approx \pm 4.9 + 3\).
Launch
Remind students that we have seen that some quadratic equations have two solutions. Take the equation \(x^2 = 25\), for example. For the equation to be true, \(x\) can be 5 and -5 because \(5^2 = 25\) and \((\text-5)^2=25\). We have been writing the solutions as: \(x = 5\) and \(x=\text-5\). Explain that a shorter way to convey the same information is by writing \(x = \pm 5\).
Student Facing
Solve each equation. Use the \(\pm\) notation when appropriate.
- \(x^2 - 13 = \text-12\)
- \((x-6)^2 = 0\)
- \(x^2 + 9 = 0\)
- \(x^2 = 18\)
- \(x^2 + 1 =18\)
- \((x + 1)^2 = 18\)
Student Response
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Anticipated Misconceptions
When using the \(\pm\) notation for the first time, some students may struggle to remember that it shows a number and its opposite. Encourage those students to continue writing their answers both ways in this lesson to reinforce the meaning of the new notation.
Some students may still struggle to understand the meaning of the square root. Watch for students who evaluate a square root by dividing by 2. Ask these students what the square root of 9 is. If this does not convince students that dividing by two is the incorrect operation, demonstrate that we call 3 a square root of 9 because \(3\boldcdot3=9\).
Activity Synthesis
Invite students to share their solutions. Record and display the solutions for all to see. Discuss any disagreements, if there are any.
Draw students’ attention to the last three sets of solutions, which are irrational. Ask students to recall the meaning of rational numbers and irrational numbers. Remind students that a rational number is a fraction or its opposite, for example: 12, -7, \(\frac18\), \(\text- \frac18\), 90.38, or 0.005. Irrational numbers are numbers that are not rational, for example: \(\sqrt2\), \( \pi,\) and \(\sqrt{10}\). (Students will have more opportunities to review and classify rational and irrational numbers later in the unit, so for now, it suffices that they remember that an irrational number is a number that is not rational.)
Highlight that when the solution is irrational, the most concise way to write an exact solution is, for example, \(x = \pm \sqrt{18} - 1\). Writing this in decimal form only allows us to write an approximate solution. For example, in this case, 3.24 and -5.24 are approximate solutions.
Then, display the variations in writing the solutions to \((x + 1)^2 = 18\) for all to see. Discuss how they are all different ways of writing the same two solutions.
- \(x= \pm \sqrt{18} - 1\)
- \(x = \sqrt{18} - 1\) and \(x = \text- \sqrt{18} - 1\)
- \(x = \text-1 \pm \sqrt{18}\)
- \(x = \text-1 + \sqrt{18}\) and \(x = \text-1 - \sqrt{18}\)
Design Principle(s): Optimize output (for explanation); Maximize meta-awareness
15.3: Finding Irrational Solutions by Completing the Square (15 minutes)
Activity
This activity allows students to integrate several skills and ideas they have learned so far: solving quadratic equations by graphing and by completing the square, using \(\pm\) notation, and expressing solutions both approximately and exactly (using the square root symbol). In solving the equations algebraically and using the notations to express and verify solutions, students practice attending to precision (MP6).
The last equation may be challenging for some students as it involves fractions and messier-looking solutions. It is not essential that all students get to this equation. They will have a chance to encounter more complicated equations and solutions in upcoming lessons.
Launch
Display the equation \(x^2 + 6x +7=0\) for all to see. Ask students if they could solve it by rewriting it in factored form. (The expression cannot be rewritten in factored form.) Remind students that they know how to solve any equation by completing the square, which would give exact solutions. They also know how to solve by graphing, which would give approximate solutions.
Arrange students in groups of two and provide access to graphing technology. One partner should find exact solutions by completing the square and the other should find approximate solutions by graphing. Partners should confirm that the \(x\)-intercepts of the graph (or the zeros of the function represented by the graph) approximate the exact solutions obtained algebraically. Ask partners to switch roles for each equation.
Design Principle(s): Support sense-making; Maximize meta-awareness
Supports accessibility for: Attention; Social-emotional skills
Student Facing
Here is an example of an equation being solved by graphing and by completing the square.
\(\displaystyle \begin {align} x^2 + 6x +7 &=0\\ x^2 + 6x + 9 &= 2\\(x+3)^2 &= 2\\x+3 &=\pm \sqrt2\\ x &=\text-3\pm \sqrt2 \end{align}\)
Verify: \(\sqrt2\) is approximately 1.414. So \(\text-3+\sqrt2 \approx \text-1.586\) and \(\text-3-\sqrt2 \approx \text-4.414\).
For each equation, find the exact solutions by completing the square and the approximate solutions by graphing. Then, verify that the solutions found using the two methods are close. If you get stuck, study the example.
- \(x^2+4x+1=0\)
- \(x^2-10x+18=0\)
- \(x^2+5x+\frac14=0\)
- \(x^2+\frac83 x + \frac{14}{9}=0\)
Student Response
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Student Facing
Are you ready for more?
Write a quadratic equation of the form \(ax^2 + bx + c = 0\) whose solutions are \(x = 5-\sqrt{2}\) and \(x = 5+\sqrt{2}\).
Student Response
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Activity Synthesis
Discuss any common struggles or common mistakes made when solving the equations. Then, invite students to reflect on the merits and challenges of solving by each method. Ask questions such as:
- “What are some benefits of solving by graphing? What are some drawbacks?” (Benefits: It is quick and straightforward. It shows the solutions, or that there are no solutions, even when the equations involve fractions or very large numbers. Drawbacks: It does not give exact solutions. Some tools may require adjusting the graphing window quite a bit to see the solutions.)
- “What are some benefits and drawbacks of solving by completing the square?” (Benefit: It can be used to find exact solutions to any equation. Drawbacks: It can be pretty time consuming. When the equations have fractions or very large or very small numbers, the calculations get complicated and may be prone to error.)
Lesson Synthesis
Lesson Synthesis
To highlight and help students connect the key ideas in the lesson, discuss questions such as:
- “How is the \(\pm\) notation useful when solving quadratic equations? Will all solutions need this notation?” (The \(\pm\) notation is useful for expressing solutions that are the two square roots of a number—for example, the solutions to \(x^2=81\). It is not needed when an equation has no solutions—for example, \(x^2=\text-5\), or only one solution—for example, \(x^2=0\).)
- “How is the \(\sqrt{\phantom{3}}\) symbol useful when solving quadratic equations?” (It helps us express the solutions exactly, especially when the solutions are irrational. We can just say, for example, that the solutions are \(\pm \sqrt{18}\), instead of using the decimal approximations.)
- “Does the expression \(\sqrt{40}\) mean there are two solutions, one positive and one negative?” (No. The \(\sqrt{\phantom{3}}\) symbol by convention refers only to the positive square root of a number. To show the negative square root, we need to write \(\text- \sqrt{40}\). Or to show both solutions, we need to write \(\pm \sqrt {40}\).)
- “What are some benefits and drawbacks of expressing the solutions using the square root symbol and the plus-minus notation—as in \(\text-3\pm \sqrt2\)—instead of writing them as decimals?” (Benefits: The solutions are exact. The radical and \(\pm\) notations are very succinct and save computation time. Drawbacks: It is not always easy to tell what the approximate values of the solutions are. When the solutions are written as -1.586 and -4.414, we have an intuition about their size or where they are on the number line. When expressed as \(\text-3\pm \sqrt2\), it isn’t immediately clear how large they are or if they are positive or negative.)
15.4: Cool-down - Finding Exact Solutions (5 minutes)
Cool-Down
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Student Lesson Summary
Student Facing
When solving quadratic equations, it is important to remember that:
- Any positive number has two square roots, one positive and one negative, because there are two numbers that can be squared to make that number. (For example, \(6^2\) and \((\text-6)^2\) both equal 36, so 6 and -6 are both square roots of 36.)
- The square root symbol (\(\sqrt{\phantom{3}}\)) can be used to express the positive square root of a number. For example, the square root of 36 is 6, but it can also be written as \(\sqrt{36}\) because \(\sqrt{36} \boldcdot \sqrt{36} = 36\).
- To express the negative square root of a number, say 36, we can write -6 or \(\text- \sqrt {36}\).
- When a number is not a perfect square—for example, 40—we can express its square roots by writing \(\sqrt{40}\) and \(\text- \sqrt{40}\).
How could we write the solutions to an equation like \((x + 4)^2 = 11\)? This equation is saying, “something squared is 11.” To make the equation true, that something must be \(\sqrt{11}\) or \(\text-\sqrt{11}\). We can write:
\(\displaystyle \begin {align} x+4 = \sqrt{11} \quad &\text{or} \quad x+4 = \text- \sqrt{11}\\x = \text-4 + \sqrt{11} \quad &\text{or} \quad x=\text-4 - \sqrt{11} \end {align}\)
A more compact way to write the two solutions to the equation is: \(x=\text-4 \pm \sqrt{11}\).
About how large or small are those numbers? Are they positive or negative? We can use a calculator to compute the approximate values of both expressions:
\(\displaystyle \text-4 + \sqrt{11} \approx \text-0.683 \quad \text{or} \quad \text-4 - \sqrt{11} \approx \text-7.317\)
We can also approximate the solutions by graphing. The equation \((x+4)^2=11\) is equivalent to \((x+4)^2-11=0\), so we can graph the function \(y=(x+4)^2-11\) and find its zeros by locating the \(x\)-intercepts of the graph.
