Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8•7
Lesson 5: Solving Equations with Radicals
Student Outcomes

Students find the positive solutions to equations algebraically equivalent to equations of the form 𝑥 2 = 𝑝 and
𝑥 3 = 𝑝.
Lesson Notes
As with previous lessons, students are asked to find only the positive value of 𝑥 that makes each equation true.
However, we have included in the examples and exercises both the positive and the negative values of 𝑥 so that the
teacher can choose whether to provide the rationale for both solutions and whether to require students to give them as
part of their answers.
Classwork
Discussion (15 minutes)

Just recently, we began solving equations that required us to find the square root or cube root of a number.
All of those equations were in the form of 𝑥 2 = 𝑝 or 𝑥 3 = 𝑝, where 𝑝 was a positive rational number.
Example 1
Example 1
𝒙𝟑 + 𝟗𝒙 =

𝟏
(𝟏𝟖𝒙 + 𝟓𝟒)
𝟐
Now that we know about square roots and cube roots, we can combine that knowledge with our knowledge of
1
2
the properties of equality to begin solving nonlinear equations like 𝑥 3 + 9𝑥 = (18𝑥 + 54). Transform the
MP.1
equation until you can determine the positive value of 𝑥 that makes the equation true.
Challenge students to solve the equation independently or in pairs. Have students share their strategy for solving the
equation. Ask them to explain each step.

Sample response:
1
(18𝑥 + 54)
2
𝑥 3 + 9𝑥 = 9𝑥 + 27
𝑥 3 + 9𝑥 =
𝑥 3 + 9𝑥 − 9𝑥 = 9𝑥 − 9𝑥 + 27
𝑥 3 = 27
Scaffolding:
Consider using a simpler
version of the equation (line 2,
for example):
𝑥 3 + 9𝑥 = 9𝑥 + 27.
3
√𝑥 3 = 3√27
3
𝑥 = √ 33
𝑥=3
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM

8•7
Now, we verify our solution is correct.
1
33 + 9(3) = (18(3) + 54)
2
1
27 + 27 = (54 + 54)
2
1
54 = (108)
2
54 = 54

Since the left side is the same as the right side, our solution is correct.
Example 2
Example 2
𝒙(𝒙 − 𝟑) − 𝟓𝟏 = −𝟑𝒙 + 𝟏𝟑

Let’s look at another nonlinear equation. Find the positive value of 𝑥 that makes the equation true:
𝑥(𝑥 − 3) − 51 = −3𝑥 + 13.
Provide students with time to solve the equation independently or in pairs. Have students share their strategy for
solving the equation. Ask them to explain each step.

Sample response:
𝑥(𝑥 − 3) − 51 = −3𝑥 + 13
𝑥 2 − 3𝑥 − 51 = −3𝑥 + 13
𝑥 2 − 3𝑥 + 3𝑥 − 51 = −3𝑥 + 3𝑥 + 13
𝑥 2 − 51 = 13
𝑥 2 − 51 + 51 = 13 + 51
𝑥 2 = 64
√𝑥 2 = ±√64
𝑥 = ±√64
𝑥 = ±8

Now we verify our solution is correct.
Provide students time to check their work.

Let 𝑥 = 8.
Let 𝑥 = −8.
8(8 − 3) − 51 = −3(8) + 13
8(5) − 51 = −24 + 13

−8(−8 − 3) − 51 = −3(−8) + 13
−8(−11) − 51 = 24 + 13
40 − 51 = −11
88 − 51 = 37
−11 = −11
37 = 37
Now it is clear that the left side is exactly the same as the right side, and our solution is correct.
Lesson 5:
Solving Equations with Radicals
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8•7
Exercises (20 minutes)
Students complete Exercises 1–7 independently or in pairs. Although we are asking students to find the positive value of
𝑥 that makes each equation true, we have included in the exercises both the positive and the negative values of 𝑥 so
that the teacher can choose whether to use them.
Exercises
Find the positive value of 𝒙 that makes each equation true, and then verify your solution is correct.
1.
a.
Solve 𝒙𝟐 − 𝟏𝟒 = 𝟓𝒙 + 𝟔𝟕 − 𝟓𝒙.
𝒙𝟐 − 𝟏𝟒 = 𝟓𝒙 + 𝟔𝟕 − 𝟓𝒙
Check:
𝒙𝟐 − 𝟏𝟒 = 𝟔𝟕
𝟗𝟐 − 𝟏𝟒 = 𝟓(𝟗) + 𝟔𝟕 − 𝟓(𝟗)
𝟖𝟏 − 𝟏𝟒 = 𝟒𝟓 + 𝟔𝟕 − 𝟒𝟓
𝟔𝟕 = 𝟔𝟕
𝒙𝟐 − 𝟏𝟒 + 𝟏𝟒 = 𝟔𝟕 + 𝟏𝟒
𝒙𝟐 = 𝟖𝟏
√𝒙𝟐 = ±√𝟖𝟏
𝒙 = ±√𝟖𝟏
𝒙 = ±𝟗
b.
(−𝟗)𝟐 − 𝟏𝟒 = 𝟓(−𝟗) + 𝟔𝟕 − 𝟓(−𝟗)
𝟖𝟏 − 𝟏𝟒 = −𝟒𝟓 + 𝟔𝟕 + 𝟒𝟓
𝟔𝟕 = 𝟔𝟕
Explain how you solved the equation.
To solve the equation, I had to first use the properties of equality to transform the equation into the form of
𝒙𝟐 = 𝟖𝟏. Then, I had to take the square root of both sides of the equation to determine that 𝒙 = 𝟗 since the
number 𝒙 is being squared.
2.
Solve and simplify: 𝒙(𝒙 − 𝟏) = 𝟏𝟐𝟏 − 𝒙.
𝒙(𝒙 − 𝟏) = 𝟏𝟐𝟏 − 𝒙
Check:
𝒙𝟐 − 𝒙 = 𝟏𝟐𝟏 − 𝒙
𝒙𝟐 − 𝒙 + 𝒙 = 𝟏𝟐𝟏 − 𝒙 + 𝒙
𝒙𝟐 = 𝟏𝟐𝟏
𝟏𝟏(𝟏𝟏 − 𝟏) = 𝟏𝟐𝟏 − 𝟏𝟏
𝟏𝟏(𝟏𝟎) = 𝟏𝟏𝟎
𝟏𝟏𝟎 = 𝟏𝟏𝟎
√𝒙𝟐 = ±√𝟏𝟐𝟏
𝒙 = ±√𝟏𝟐𝟏
𝒙 = ±𝟏𝟏
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G8-M7-TE-1.3.0-10.2015
−𝟏𝟏(−𝟏𝟏 − 𝟏) = 𝟏𝟐𝟏 − (−𝟏𝟏)
−𝟏𝟏(−𝟏𝟐) = 𝟏𝟐𝟏 + 𝟏𝟏
𝟏𝟑𝟐 = 𝟏𝟑𝟐
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
3.
8•7
A square has a side length of 𝟑𝒙 inches and an area of 𝟑𝟐𝟒 𝐢𝐧𝟐 . What is the value of 𝒙?
(𝟑𝒙)𝟐 = 𝟑𝟐𝟒
Check:
𝟑𝟐 𝒙𝟐 = 𝟑𝟐𝟒
(𝟑(𝟔)) 𝟐 = 𝟑𝟐𝟒
𝟗𝒙𝟐 = 𝟑𝟐𝟒
𝟏𝟖𝟐 = 𝟑𝟐𝟒
𝟑𝟐𝟒 = 𝟑𝟐𝟒
𝟗𝒙𝟐 𝟑𝟐𝟒
=
𝟗
𝟗
𝟐
𝒙 = 𝟑𝟔
√𝒙𝟐 = √𝟑𝟔
𝒙=𝟔
A negative number would not make sense as a length, so 𝒙 = 𝟔.
4.
−𝟑𝒙𝟑 + 𝟏𝟒 = −𝟔𝟕
−𝟑𝒙𝟑 + 𝟏𝟒 = −𝟔𝟕
Check:
−𝟑𝒙𝟑 + 𝟏𝟒 − 𝟏𝟒 = −𝟔𝟕 − 𝟏𝟒
−𝟑(𝟑)𝟑 + 𝟏𝟒 = −𝟔𝟕
𝟑
−𝟑𝒙 = −𝟖𝟏
−𝟑(𝟐𝟕) + 𝟏𝟒 = −𝟔𝟕
−𝟑𝒙𝟑 −𝟖𝟏
=
−𝟑
−𝟑
𝒙𝟑 = 𝟐𝟕
−𝟖𝟏 + 𝟏𝟒 = −𝟔𝟕
−𝟔𝟕 = −𝟔𝟕
𝟑
√𝒙𝟑 = 𝟑√𝟐𝟕
𝒙=𝟑
5.
𝒙(𝒙 + 𝟒) − 𝟑 = 𝟒(𝒙 + 𝟏𝟗. 𝟓)
𝒙(𝒙 + 𝟒) − 𝟑 = 𝟒(𝒙 + 𝟏𝟗. 𝟓)
Check:
𝒙𝟐 + 𝟒𝒙 − 𝟑 = 𝟒𝒙 + 𝟕𝟖
𝟗(𝟗 + 𝟒) − 𝟑 = 𝟒(𝟗 + 𝟏𝟗. 𝟓)
𝒙𝟐 + 𝟒𝒙 − 𝟒𝒙 − 𝟑 = 𝟒𝒙 − 𝟒𝒙 + 𝟕𝟖
𝟗(𝟏𝟑) − 𝟑 = 𝟒(𝟐𝟖. 𝟓)
𝒙𝟐 − 𝟑 = 𝟕𝟖
𝟏𝟏𝟕 − 𝟑 = 𝟏𝟏𝟒
𝒙𝟐 − 𝟑 + 𝟑 = 𝟕𝟖 + 𝟑
𝟏𝟏𝟒 = 𝟏𝟏𝟒
𝒙𝟐 = 𝟖𝟏
√𝒙𝟐 = ±√𝟖𝟏
−𝟗(−𝟗 + 𝟒) − 𝟑 = 𝟒(−𝟗 + 𝟏𝟗. 𝟓)
𝒙 = ±𝟗
−𝟗(−𝟓) − 𝟑 = 𝟒(𝟏𝟎. 𝟓)
𝟒𝟓 − 𝟑 = 𝟒𝟐
𝟒𝟐 = 𝟒𝟐
6.
𝟐𝟏𝟔 + 𝒙 = 𝒙(𝒙𝟐 − 𝟓) + 𝟔𝒙
𝟐𝟏𝟔 + 𝒙 = 𝒙(𝒙𝟐 − 𝟓) + 𝟔𝒙
Check:
𝟐𝟏𝟔 + 𝒙 = 𝒙𝟑 − 𝟓𝒙 + 𝟔𝒙
𝟐𝟏𝟔 + 𝒙 = 𝒙𝟑 + 𝒙
𝟐𝟏𝟔 + 𝒙 − 𝒙 = 𝒙𝟑 + 𝒙 − 𝒙
𝟐𝟏𝟔 = 𝒙𝟑
𝟑
𝟐𝟏𝟔 + 𝟔 = 𝟔(𝟔𝟐 − 𝟓) + 𝟔(𝟔)
𝟐𝟐𝟐 = 𝟔(𝟑𝟏) + 𝟑𝟔
𝟐𝟐𝟐 = 𝟏𝟖𝟔 + 𝟑𝟔
𝟐𝟐𝟐 = 𝟐𝟐𝟐
𝟑
√𝟐𝟏𝟔 = √𝒙𝟑
𝟔=𝒙
Lesson 5:
Solving Equations with Radicals
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8•7
7.
a.
What are we trying to determine in the diagram below?
We need to determine the value of 𝒙 so that its square root, multiplied by 𝟒,
satisfies the equation
𝟐
𝟓𝟐 + (𝟒√𝒙) = 𝟏𝟏𝟐.
b.
Determine the value of 𝒙, and check your answer.
𝟐
𝟓𝟐 + (𝟒√𝒙) = 𝟏𝟏𝟐
Check:
𝟐
𝟐𝟓 + 𝟒𝟐 (√𝒙) = 𝟏𝟐𝟏
𝟐
𝟐𝟓 − 𝟐𝟓 + 𝟒𝟐 (√𝒙) = 𝟏𝟐𝟏 − 𝟐𝟓
𝟐
𝟓𝟐 + (𝟒√𝟔) = 𝟏𝟏𝟐
𝟐𝟓 + 𝟏𝟔(𝟔) = 𝟏𝟐𝟏
𝟏𝟔𝒙 = 𝟗𝟔
𝟏𝟔𝒙 𝟗𝟔
=
𝟏𝟔
𝟏𝟔
𝒙=𝟔
𝟐𝟓 + 𝟗𝟔 = 𝟏𝟐𝟏
𝟏𝟐𝟏 = 𝟏𝟐𝟏
The value of 𝒙 is 𝟔.
Closing (5 minutes)
Summarize, or ask students to summarize, the main points from the lesson:

We know how to solve equations with squared and cubed variables and verify that our solutions are correct.
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G8-M7-TE-1.3.0-10.2015
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8•7
Lesson Summary
Equations that contain variables that are squared or cubed can be solved using the properties of equality and the
definition of square and cube roots.
Simplify an equation until it is in the form of 𝒙𝟐 = 𝒑 or 𝒙𝟑 = 𝒑, where 𝒑 is a positive rational number; then, take the
square or cube root to determine the positive value of 𝒙.
Example:
Solve for 𝒙.
𝟏
(𝟐𝒙𝟐 + 𝟏𝟎) = 𝟑𝟎
𝟐
𝒙𝟐 + 𝟓 = 𝟑𝟎
Check:
𝒙𝟐 + 𝟓 − 𝟓 = 𝟑𝟎 − 𝟓
𝒙𝟐 = 𝟐𝟓
√𝒙𝟐 = √𝟐𝟓
𝒙=𝟓
𝟏
(𝟐(𝟓)𝟐 + 𝟏𝟎) = 𝟑𝟎
𝟐
𝟏
(𝟐(𝟐𝟓) + 𝟏𝟎) = 𝟑𝟎
𝟐
𝟏
(𝟓𝟎 + 𝟏𝟎) = 𝟑𝟎
𝟐
𝟏
(𝟔𝟎) = 𝟑𝟎
𝟐
𝟑𝟎 = 𝟑𝟎
Exit Ticket (5 minutes)
Lesson 5:
Solving Equations with Radicals
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
Name
8•7
Date
Lesson 5: Solving Equations with Radicals
Exit Ticket
1.
Find the positive value of 𝑥 that makes the equation true, and then verify your solution is correct.
𝑥 2 + 4𝑥 = 4(𝑥 + 16)
2.
Find the positive value of 𝑥 that makes the equation true, and then verify your solution is correct.
(4𝑥)3 = 1728
Lesson 5:
Solving Equations with Radicals
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This file derived from G8-M7-TE-1.3.0-10.2015
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
8•7
Exit Ticket Sample Solutions
1.
Find the positive value of 𝒙 that makes the equation true, and then verify your solution is correct.
𝒙𝟐 + 𝟒𝒙 = 𝟒(𝒙 + 𝟏𝟔)
𝒙𝟐 + 𝟒𝒙 = 𝟒(𝒙 + 𝟏𝟔)
Check:
𝟐
𝒙 + 𝟒𝒙 = 𝟒𝒙 + 𝟔𝟒
𝟖𝟐 + 𝟒(𝟖) = 𝟒(𝟖 + 𝟏𝟔)
𝒙𝟐 + 𝟒𝒙 − 𝟒𝒙 = 𝟒𝒙 − 𝟒𝒙 + 𝟔𝟒
𝟔𝟒 + 𝟑𝟐 = 𝟒(𝟐𝟒)
𝒙𝟐 = 𝟔𝟒
𝟗𝟔 = 𝟗𝟔
√𝒙𝟐 = √𝟔𝟒
𝒙=𝟖
2.
Find the positive value of 𝒙 that makes the equation true, and then verify your solution is correct.
(𝟒𝒙)𝟑 = 𝟏𝟕𝟐𝟖
(𝟒𝒙)𝟑 = 𝟏𝟕𝟐𝟖
Check:
𝟑
𝟔𝟒𝒙 = 𝟏𝟕𝟐𝟖
𝟏
𝟏
(𝟔𝟒𝒙𝟑 ) = (𝟏𝟕𝟐𝟖)
𝟔𝟒
𝟔𝟒
𝒙𝟑 = 𝟐𝟕
𝟑
√ 𝒙𝟑
𝟑
(𝟒(𝟑)) = 𝟏𝟕𝟐𝟖
𝟏𝟐𝟑 = 𝟏𝟕𝟐𝟖
𝟏𝟕𝟐𝟖 = 𝟏𝟕𝟐𝟖
𝟑
= √𝟐𝟕
𝒙=𝟑
Problem Set Sample Solutions
Find the positive value of 𝒙 that makes each equation true, and then verify your solution is correct.
1.
𝟏
𝟐
𝒙𝟐 (𝒙 + 𝟕) = (𝟏𝟒𝒙𝟐 + 𝟏𝟔)
𝟏
(𝟏𝟒𝒙𝟐 + 𝟏𝟔)
𝟐
𝒙𝟑 + 𝟕𝒙𝟐 = 𝟕𝒙𝟐 + 𝟖
𝒙𝟐 (𝒙 + 𝟕) =
Check:
𝒙𝟑 + 𝟕𝒙𝟐 − 𝟕𝒙𝟐 = 𝟕𝒙𝟐 − 𝟕𝒙𝟐 + 𝟖
𝒙𝟑 = 𝟖
𝟑
√𝒙𝟑 = 𝟑√𝟖
𝒙=𝟐
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G8-M7-TE-1.3.0-10.2015
𝟏
(𝟏𝟒(𝟐𝟐 ) + 𝟏𝟔)
𝟐
𝟏
𝟒(𝟗) = (𝟓𝟔 + 𝟏𝟔)
𝟐
𝟏
𝟑𝟔 = (𝟕𝟐)
𝟐
𝟑𝟔 = 𝟑𝟔
𝟐𝟐 (𝟐 + 𝟕) =
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
2.
𝒙𝟑 = 𝟏𝟑𝟑𝟏−𝟏
𝒙𝟑 = 𝟏𝟑𝟑𝟏−𝟏
𝟑
√ 𝒙𝟑
=
√𝟏𝟑𝟑𝟏−𝟏
𝟑
𝟏
𝟏𝟑𝟑𝟏
𝟑
𝟏
𝟏𝟏𝟑
𝒙=√
𝒙=
Check:
𝟑
𝒙=√
3.
8•7
𝟏 𝟑
( ) = 𝟏𝟑𝟑𝟏−𝟏
𝟏𝟏
𝟏
= 𝟏𝟑𝟑𝟏−𝟏
𝟏𝟏𝟑
𝟏
= 𝟏𝟑𝟑𝟏−𝟏
𝟏𝟑𝟑𝟏
𝟏𝟑𝟑𝟏−𝟏 = 𝟏𝟑𝟑𝟏−𝟏
𝟏
𝟏𝟏
Determine the positive value of 𝒙 that makes the equation true, and then explain how you solved the equation.
𝒙𝟗
𝒙𝟕
− 𝟒𝟗 = 𝟎
𝒙𝟗
− 𝟒𝟗 = 𝟎
𝒙𝟕
𝒙𝟐 − 𝟒𝟗 = 𝟎
Check:
𝟕𝟐 − 𝟒𝟗 = 𝟎
𝟒𝟗 − 𝟒𝟗 = 𝟎
𝒙𝟐 − 𝟒𝟗 + 𝟒𝟗 = 𝟎 + 𝟒𝟗
𝟎=𝟎
𝒙𝟐 = 𝟒𝟗
√𝒙𝟐 = √𝟒𝟗
𝒙=𝟕
To solve the equation, I first had to simplify the expression
𝒙𝟗
𝒙𝟕
to 𝒙𝟐 . Next, I used the properties of equality to
transform the equation into 𝒙𝟐 = 𝟒𝟗. Finally, I had to take the square root of both sides of the equation to solve for
𝒙.
4.
Determine the positive value of 𝒙 that makes the equation true.
(𝟖𝒙)𝟐 = 𝟏
(𝟖𝒙)𝟐 = 𝟏
Check:
𝟔𝟒𝒙𝟐 = 𝟏
√𝟔𝟒𝒙𝟐 = √𝟏
𝟖𝒙 = 𝟏
𝟖𝒙 𝟏
=
𝟖
𝟖
𝟏
𝒙=
𝟖
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G8-M7-TE-1.3.0-10.2015
𝟐
𝟏
(𝟖 ( )) = 𝟏
𝟖
𝟏𝟐 = 𝟏
𝟏=𝟏
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
5.
𝟐
(𝟗√𝒙) − 𝟒𝟑𝒙 = 𝟕𝟔
𝟐
(𝟗√𝒙) − 𝟒𝟑𝒙 = 𝟕𝟔
Check:
𝟐
𝟐
𝟗𝟐 (√𝒙) − 𝟒𝟑𝒙 = 𝟕𝟔
(𝟗(√𝟐)) − 𝟒𝟑(𝟐) = 𝟕𝟔
𝟖𝟏𝒙 − 𝟒𝟑𝒙 = 𝟕𝟔
𝟐
𝟗𝟐 (√𝟐) − 𝟖𝟔 = 𝟕𝟔
𝟑𝟖𝒙 = 𝟕𝟔
𝟑𝟖𝒙 𝟕𝟔
=
𝟑𝟖
𝟑𝟖
𝒙=𝟐
6.
8•7
𝟖𝟏(𝟐) − 𝟖𝟔 = 𝟕𝟔
𝟏𝟔𝟐 − 𝟖𝟔 = 𝟕𝟔
𝟕𝟔 = 𝟕𝟔
Determine the length of the hypotenuse of the right triangle below.
𝟑𝟐 + 𝟕𝟐 = 𝒙𝟐
Check:
𝟗 + 𝟒𝟗 = 𝒙𝟐
𝟓𝟖 = 𝒙
𝟐
√𝟓𝟖 = √𝒙𝟐
𝟐
𝟑𝟐 + 𝟕𝟐 = √𝟓𝟖
𝟗 + 𝟒𝟗 = 𝟓𝟖
𝟓𝟖 = 𝟓𝟖
√𝟓𝟖 = 𝒙
Since 𝒙 = √𝟓𝟖 , the length of the hypotenuse is √𝟓𝟖 𝐦𝐦.
Lesson 5:
Solving Equations with Radicals
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
7.
8•7
Determine the length of the legs in the right triangle below.
𝟐
Check:
𝒙𝟐 + 𝒙𝟐 = (𝟏𝟒√𝟐)
𝟐
𝟐
𝟐𝒙𝟐 = 𝟏𝟒𝟐 (√𝟐)
𝟏𝟒𝟐 + 𝟏𝟒𝟐 = (𝟏𝟒√𝟐)
𝟐𝒙𝟐 = 𝟏𝟗𝟔(𝟐)
𝟏𝟗𝟔 + 𝟏𝟗𝟔 = 𝟏𝟒𝟐 (√𝟐)
𝟐
𝟐
𝟐𝒙
𝟏𝟗𝟔(𝟐)
=
𝟐
𝟐
𝒙𝟐 = 𝟏𝟗𝟔
𝟑𝟗𝟐 = 𝟏𝟗𝟔(𝟐)
𝟑𝟗𝟐 = 𝟑𝟗𝟐
√𝒙𝟐 = √𝟏𝟗𝟔
𝒙 = √𝟏𝟒𝟐
𝒙 = 𝟏𝟒
Since 𝒙 = 𝟏𝟒, the length of each of the legs of the right triangle is 𝟏𝟒 𝐜𝐦.
8.
An equilateral triangle has side lengths of 𝟔 𝐜𝐦. What is the height of the triangle? What is the area of the
triangle?
Note: This problem has two solutions, one with a simplified root and one without. Choose the appropriate solution
for your classes based on how much simplifying you have taught them.
Let 𝒉 𝐜𝐦 represent the height of the triangle.
𝟑𝟐 + 𝒉𝟐 = 𝟔𝟐
𝟗 + 𝒉𝟐 = 𝟑𝟔
𝟗 − 𝟗 + 𝒉𝟐 = 𝟑𝟔 − 𝟗
𝒉𝟐 = 𝟐𝟕
√𝒉𝟐 = √𝟐𝟕
𝒉 = √𝟐𝟕
𝒉 = √𝟑𝟑
𝒉 = √𝟑𝟐 × √𝟑
𝒉 = 𝟑√𝟑
Let 𝑨 represent the area of the triangle.
𝑨=
𝟔(𝟑√𝟑)
𝟐
𝑨 = 𝟑(𝟑√𝟑)
𝑨 = 𝟗√𝟑
Simplified: The height of the triangle is 𝟑√𝟑 𝐜𝐦, and the area is 𝟗√𝟑 𝐜𝐦𝟐 .
Unsimplified: The height of the triangle is √𝟐𝟕 𝐜𝐦, and the area is 𝟑√𝟐𝟕 𝐜𝐦𝟐.
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G8-M7-TE-1.3.0-10.2015
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Lesson 5
NYS COMMON CORE MATHEMATICS CURRICULUM
9.
8•7
Challenge: Find the positive value of 𝒙 that makes the equation true.
𝟏
𝟐
𝟐
( 𝒙) − 𝟑𝒙 = 𝟕𝒙 + 𝟖 − 𝟏𝟎𝒙
𝟏 𝟐
( 𝒙) − 𝟑𝒙 = 𝟕𝒙 + 𝟖 − 𝟏𝟎𝒙
𝟐
𝟏 𝟐
𝒙 − 𝟑𝒙 = −𝟑𝒙 + 𝟖
𝟒
Check:
𝟐
𝟏
( (𝟒√𝟐)) − 𝟑(𝟒√𝟐) = 𝟕(𝟒√𝟐) + 𝟖 − 𝟏𝟎(𝟒√𝟐)
𝟐
𝟏
(𝟏𝟔)(𝟐) − 𝟑(𝟒√𝟐) = 𝟕(𝟒√𝟐) − 𝟏𝟎(𝟒√𝟐) + 𝟖
𝟒
𝟑𝟐
− 𝟑(𝟒√𝟐) = 𝟕(𝟒√𝟐) − 𝟏𝟎(𝟒√𝟐) + 𝟖
𝟒
𝟖 − 𝟑(𝟒√𝟐) = (𝟕 − 𝟏𝟎)(𝟒√𝟐) + 𝟖
𝟏 𝟐
𝒙 − 𝟑𝒙 + 𝟑𝒙 = −𝟑𝒙 + 𝟑𝒙 + 𝟖
𝟒
𝟏 𝟐
𝒙 =𝟖
𝟒
𝟏 𝟐
𝟒 ( ) 𝒙 = 𝟖(𝟒)
𝟒
𝒙𝟐 = 𝟑𝟐
𝟖 − 𝟑(𝟒√𝟐) = −𝟑(𝟒√𝟐) + 𝟖
𝟖 − 𝟖 − 𝟑(𝟒√𝟐) = −𝟑(𝟒√𝟐) + 𝟖 − 𝟖
√𝒙𝟐 = √𝟑𝟐
−𝟑(𝟒√𝟐) = −𝟑(𝟒√𝟐)
𝒙 = √𝟐𝟓
𝒙 = √𝟐𝟐 ⋅ √𝟐𝟐 ⋅ √𝟐
𝒙 = 𝟒√𝟐
10. Challenge: Find the positive value of 𝒙 that makes the equation true.
𝟏𝟏𝒙 + 𝒙(𝒙 − 𝟒) = 𝟕(𝒙 + 𝟗)
𝟏𝟏𝒙 + 𝒙(𝒙 − 𝟒) = 𝟕(𝒙 + 𝟗)
Check:
𝟏𝟏𝒙 + 𝒙𝟐 − 𝟒𝒙 = 𝟕𝒙 + 𝟔𝟑
𝟕𝒙 + 𝒙𝟐 = 𝟕𝒙 + 𝟔𝟑
𝟕𝒙 − 𝟕𝒙 + 𝒙𝟐 = 𝟕𝒙 − 𝟕𝒙 + 𝟔𝟑
𝟏𝟏(𝟑√𝟕) + 𝟑√𝟕(𝟑√𝟕 − 𝟒) = 𝟕(𝟑√𝟕 + 𝟗)
𝟐
𝟑𝟑√𝟕 + 𝟑𝟐 (√𝟕) − 𝟒(𝟑√𝟕) = 𝟐𝟏√𝟕 + 𝟔𝟑
𝒙𝟐 = 𝟔𝟑
√𝒙𝟐 = √𝟔𝟑
𝒙 = √(𝟑𝟐 )(𝟕)
𝒙 = √𝟑𝟐 ⋅ √𝟕
𝒙 = 𝟑√𝟕
𝟑𝟑√𝟕 − 𝟒(𝟑√𝟕) + 𝟗(𝟕) = 𝟐𝟏√𝟕 + 𝟔𝟑
𝟑𝟑√𝟕 − 𝟏𝟐√𝟕 + 𝟔𝟑 = 𝟐𝟏√𝟕 + 𝟔𝟑
(𝟑𝟑 − 𝟏𝟐)√𝟕 + 𝟔𝟑 = 𝟐𝟏√𝟕 + 𝟔𝟑
𝟐𝟏√𝟕 + 𝟔𝟑 = 𝟐𝟏√𝟕 + 𝟔𝟑
𝟐𝟏√𝟕 + 𝟔𝟑 − 𝟔𝟑 = 𝟐𝟏√𝟕 + 𝟔𝟑 − 𝟔𝟑
𝟐𝟏√𝟕 = 𝟐𝟏√𝟕
Lesson 5:
Solving Equations with Radicals
This work is derived from Eureka Math ™ and licensed by Great Minds. ©2015 Great Minds. eureka-math.org
This file derived from G8-M7-TE-1.3.0-10.2015
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