Part 5: Solving Equations with Radicals_L08_U02_Ready for Review

Connections 3 Lesson 8: Square and Cube Roots Part 5: Solving Equations with Radicals_L08_U02_Ready for Review

Part 5: Solving Equations with Radicals
Complete Part 5 of your Lesson 8 Interactive Notes as you work through your lesson.

Radicals are used in a wide variety of equations and expressions. In this part of the lesson, you will work through examples of the types of problems that often include radicals.

Record and solve each problem in the corresponding section of your Lesson 8 Interactive Notes. When you are finished, you will have a collection of examples that will help you solve future equations and expressions that include radicals.

Simplify the numerical expression $\sqrt{50 + 94}$

Step 1:

Simplify the terms within the radical sign.
$\sqrt{50 + 94} \sqrt{144}$

Step 2:

Simplify the square root.
144 is a perfect square.
$12 \times 12 = 144$
The $\sqrt{144}$ can be simplified to 12.

Final simplified form: 12

Evaluate the following expression.

$8 + {(5 - 2)}^{2} \times 4 + \sqrt[3]{27}$

To evaluate this expression you will need to use the Order of Operations.

Recall the Order of Operations:
1: Parentheses and Brackets
2: Exponents and Radicals
3: Multiplication and Division
4: Addition and Subtraction

Step 1:
Simplify the terms inside of the parentheses.
$8 + {(5 - 2)}^{2} \times 4 + \sqrt[3]{27} 8 + {(3)}^{2} \times 4 + \sqrt[3]{27}$

Step 2:
Working in order from left to right, simplify any terms that include an exponent or radical.
$8 + {(3)}^{2} \times 4 + \sqrt[3]{27} 8 + 9 \times 4 + \sqrt[3]{27} 8 + 9 \times 4 + 3$

Step 3:
Working in order from left to right, simplify any terms that are being multiplied or divided.
$8 + 9 \times 4 + 3 8 + 36 + 3$

Step 4:
Working in order from left to right, simplify any terms that are being added or subtracted.
$8 + 36 + 3 47$

Final Simplified Form: 47

A local market is running a special sale on produce items. Each day customers will receive a discount that is equal to the date of the month squared, beginning with one cent.
For example, on the 2nd day of the month, customers will be given a discount that is equal to $2^{2}$ or 4 cents off of the price of their items.
On which day will customers receive a discount of $0.49?

Think:
The discount of $0.49 is the product of the date multiplied by itself, the date squared.
The date will be equal to the square root of 49.

Solve.
$\sqrt{49} = x$
The variable x, or any other variable, can be used to represent the value you are solving for.

49 is a perfect square.
$7 \times 7 = 49 7^{2} = 49$

$\sqrt{49} = 7 x = 7$

Final Solution:
Customers will receive a discount of $0.49 on the seventh day of the month.

The area of a square is $400 m^{2}$ . What is the length of each side of the square?
Hint: All sides of a square are equal, so the area of a square can be found using the formula side × side = area.

Think:
The area of a square is found by multiplying one side length by another side length.
Since the sides are equal, the same number is multiplied by itself, the length of one side is squared.
The length of each side is equal to the square root of the area.

Solve.
$\sqrt{400} = x$
The variable x, or any other variable, can be used to represent the value you are solving for.

400 must be a perfect square because it is the product of two equal sides.

Determine what number produces a product of 400 when multiplied by itself.
$20 \times 20 = 400 20^{2} = 400$

$\sqrt{400} = 20 x = 20$

Final Solution:
Each side of the square is 20 m. long.

Try it out!
Evaluate each problem, use the examples above as a guide.

$1) \sqrt{3 \cdot 7 + 4}$
$2) \frac{16}{4} + \sqrt{121} \times 2^{4} - 7$
3) James created a square garden with an area of $225 f t .^{2}$ .
What is the length of each side of the garden?

**Solution text below.
Compare your solutions to those shown below.

$1) \sqrt{3 \cdot 7 + 4} \sqrt{21 + 4} \sqrt{25} 5$

$2) \frac{16}{4} + \sqrt{121} \times 2^{4} - 7 \frac{16}{4} + 11 \times 2^{4} - 7 \frac{16}{4} + 11 \times 16 - 7 4 + 11 \times 16 - 7 4 + 176 - 7 180 - 7 173$

3) James created a square garden with an area of $225 f t .^{2}$
What is the length of each side of the garden?
The length of each side is equal to the square root of the area of the garden.
This problem can be represented by the equation: $\sqrt{225} = x$ where x equals the length of each side of the garden

225 is a perfect square.
$15 \times 15 = 225 15^{2} = 225$
$\sqrt{225} = 15 x = 15$
Final solution:
Each side of James’ garden is 15 ft. long.

**Instructor Notes – Text will appear when button is clicked

How to Support Students:

The problems in this part of the lesson represent a variety of problems that students will be asked to solve involving radicals. If students need more support, identify the type of problem that they need support with. Provide more practice with this type of problem.

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