John has tutored algebra and SAT Prep and has a B.A. degree with a major in psychology and a minor in mathematics from Christopher Newport University.
Algebra I Assignment - Simplifying & Solving Square Roots & Radical Expressions
Square Roots and Radical Expressions Assignment Explanation and Topic Overview
While this topic can seem as confusing to your students as the inexplicable movie Star Trek Into Darkness, by breaking each explanation into simple, logical steps, we can convey the methodologies.
The lesson will include the four basic math operations as they relate to equations, and how to rationalize a denominator too.
By the end of this assignment, your students will have completed three steps, solved nine problems, and delivered a suitable presentation.
Note: Detailed answers are located at the bottom of the page.
For further information, visit our helpful High School Algebra I: Homeschool Curriculum.
Key Terms
- Radical: a math expression which utilizes a root
- Radicand: the number inside the radical symbol, of which you are trying to take the square root
- Square root: the opposite of squaring a number
Materials
- Internet access
- Paper
- Pencil
Time/Length
- One day to complete this Algebra I assignment
- One week to create your relevant square roots and radical expressions deliverable
Assignment Instructions for Students
Step One
Part I
First of all, how do you calculate the square root of a number? That's easy! I just push this cool-looking little calculator button. While that indeed may be true, let's learn to solve square roots just like the ancient civilizations.
By the way, square roots of perfect squares are easy to find. The square root of 1 is 1, the square root of 4 is 2, and the square root of 9 is 3. See what higher numbers you can calculate.
Problem #1:
Find the square root of 19.
Part II
Next, let's simplify square roots of numbers and expressions containing square roots.
Here's your handy formula:
- (xy)1/2 = x1/2 * y1/2
Now read this sentence out loud. ''The square root of quantity x times y is equal to the square root of x times the square root of y.''
Problem #2:
Reduce the following square root, which features 18 as your radicand, to its simplest form:
- 181/2
Part III
How do we determine the square root of quotients though?
This time our handy formula is:
- (x / y)1/2 = x1/2 / y1/2
Problem #3:
- (4/9)1/2 = ?
Part IV
What if your square root has a variable with an exponent?
Problem #4:
Simplify:
- ((63x2))1/2 = ?
Step Two
Part A
By the way, how do we find the reciprocal of a given radical expression? We simply flip our numerator and our denominator:
Problem #5:
See if you can find the reciprocal of:
- x / (x + 3)
Part B
Next on our list, let's rationalize a denominator which contains a radical expression.
Problem #6:
Suppose you are given a radical expression like:
- 1/(31/2), which reads ''one divided by the square root of three''
However, you don't want that radical to appear in your denominator. Our solution is easy and oh so clever, if I do say so myself.
Step Three
Part 1
You may recall that to add and subtract using radical notation, it is important we combine like terms whenever possible.
- 5 * 71/2 + 9 * 71/2 = 14 * 71/2
That reads as ''5 times the square root of 7 plus 9 times the square root of seven equals 14 times the square root of 7.''
- 8 * 21/2 - 5 * 21/2 = 3 * 21/2
That reads as 8 times the square root of 2 minus 5 times the square root of 2 equals 3 times the square root of 2.
Next, we multiply radical expressions with two or more terms:
Problem #7:
Rewrite:
- (3a)1/2 (21/2 + c) = ?
Part 2
Let's try a more complex expression though:
Problem #8:
Rewrite:
- (3 + a1/2) (4 + 21/2) = ?
Part 3
Next, let's actually solve a radical equation with a radical term.
Problem #9:
- (a - 5 )1/2 = 7
Deliverable
Now it's time to choose one of these three presentations to show you grasp this new algebra material:
- Write practice questions for other students to be answered on paper.
- Create your own quiz show questions, based on one of your favorite game shows, to test fellow student contestants on these concepts.
- Take charge and ''be the teacher'', as you make a podcast or video to explain how to answer these questions or explain these processes.
Answers
Solution #1:
Find the square root of 19.
Ask yourself between which two perfect squares 19 is located:
- 12 = 1
- 22 = 4
- 32 = 9
- 42 = 16
- 52 = 25
Aha! We see 19 is located between 16 and 25, which are the squares of 4 and 5, respectively, so next we divide 19 by the 4 we chose to get:
- 4.75
However, 4.75 * 4.75 = 22.56, which isn't exactly really close to 19. Let's repeat the process. First, add 4 (our original divisor) + 4.75 (our result from dividing) and then divide your sum by 2 to get:
- 4.375
Now divide 19 by this 4.375 to get:
- 4.3429
Hey, 4.3429 * 4.3429 = 18.8608, which is pretty close to 19. We're done. Let's go watch that cool new Netflix show about the zombies. Hang on there. Let's repeat the process again.
- 19 divided by 4.3429 = 4.3750
Add 4.3750 (our newest result) + 4.3429 (our previous result) and divide the sum by 2 to get:
- 4.3590
Hey, 4.3590 * 4.3590 = 19.0009. That's pretty close to 19. You can go watch those zombies now.
Check your answer with those calculator buttons:
- Square root of 19 = 4.3589, which is pretty close to the 4.3590 you just calculated. In fact, they are only .0001 (one-ten-thousandth) apart from each other.
Solution #2:
- 181/2
Find the factors of 18:
- 18 * 1
- 9 * 2
- 6 * 3
That's all I can find. If you can find another one let me know. You'll be famous and appear on Kelly Clarkson's talk show.
Now, of the above factors, we see that only 1 and 9 are perfect squares. We choose 9, and apply our formula:
- (xy)1/2 = x1/2 * y1/2
- (9 * 2)1/2 = 91/2 * 21/2
And your answer in simplest form is:
- 181/2 = 3 * 21/2
You read the equation directly above as ''the square root of 18 equals 3 times the square root of two.''
Check your answer:
- The square root of 18 = 4.2426
- Since the square root of 2 is 1.4142, 3 times the square root of 2 is 3 times 1.4142 = 4.2426
Hey! They match exactly to four decimal places.
Solution #3:
- (x / y)1/2 = x1/2 / y1/2
- (4/9)1/2 = 41/2 / 91/2
- (4/9)1/2 = 2/3
It worked. Wow! We're good at this algebra stuff today.
Solution #4:
- ((63x2))1/2 =
Factor:
- (3 * 3 * 7 * x * x)1/2 =
Rewrite as:
- (3 * 3)1/2 * 71/2 * (x * x)1/2 =
- 3x * 71/2
The equation directly above reads as ''3 times x times the square root of 7'', which we often shorten to ''3x square root of 7.''
Solution #5:
- Equation: x / (x + 3)
- Reciprocal: (x + 3) / x
Solution #6:
- 1/(31/2)
Simply multiply both the numerator and denominator by the denominator:
- 1/(31/2) * (31/2 / 31/2) =
- 31/2 / 3
See how the radical took a little vacation and ''traveled'' from the denominator to the numerator instead?
Solution #7:
This looks like a job for our old friend the distributive property, which you may remember is:
- a(b + c) = ab + ac
- (3a)1/2 (21/2 + c) =
- (6a)1/2 + c(3a)1/2
At this point, we can't add the unlike terms, so we can't reduce any further.
Solution #8:
- (3 + a1/2) (4 + 21/2)
Uh-oh! We can't use our distributive property this time, as we have a binomial times a binomial. That's okay though. We have our other old friend the FOIL method.
Do you remember FOIL stands for first, outer, inner, and last?
- First: 3 * 4 = 12
- Outer: 3 * 21/2
- Inner: 4 * a1/2
- Last: a1/2 * 21/2
Now we add the terms together:
- 12 + (3 * 21/2) + (4 * a1/2) + (a1/2 * 21/2)
Solution #9:
Take a deep breath!
- (a - 5 )1/2 = 7
Hey, let's square both sides to get rid of that annoying radical!
- ((a - 5 )1/2)2 = 72
- a - 5 = 49
- a = 54
Check your answer by plugging a = 54 back into your original equation:
- (a - 5 )1/2 = 7
- (54 - 5)1/2 = 7
- 491/2 = 7
- 7 = 7
Hey, I'm pretty sure 7 = 7. It checks out correctly.
Assignment Rubric
Requirements | 0-5 points |
---|---|
Problem #1 is correctly solved | |
Problem #2 is correctly solved | |
Problem #3 is correctly solved | |
Problem #4 is correctly solved | |
Problem #5 is correctly solved | |
Problem #6 is correctly solved | |
Problem #7 is correctly solved | |
Problem #8 is correctly solved | |
Problem #9 is correctly solved | |
Deliverable is clear and concise | |
Total: | /50 points |
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BackAlgebra I Assignment - Simplifying & Solving Square Roots & Radical Expressions
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