vocabulary - rohls.weebly.com · main idea: a radical equation is an equation that has a variable...
TRANSCRIPT
Cop
yrig
ht ©
by
Pear
son
Educ
atio
n, In
c. o
r its
aff
iliat
es. A
ll Ri
ghts
Res
erve
d.
Vocabulary
Review
Chapter 10 302
10-4 Solving Radical Equations
Draw a line from each equation in Column A to its solution in Column B.
Column A Column B
1. 3x 1 5 5 24 x 5 3
2. 4x2 5 16 x 5 22 or x 5 2
3. 6x 2 2 5 16 x 5 23
Vocabulary Builder
extraneous (adjective) ek stray nee us
Definition: Extraneous means not essential or relevant to the situation or subject being considered.
Math Usage: An extraneous solution is one that comes from the solving process but is not a valid solution of the original equation.
Main Idea: A radical equation is an equation that has a variable in a radicand. When solving radical equations, it is possible to obtain extraneous solutions. You must check each solution in the original equation.
Use Your Vocabulary
If you solve the equation 2!x 5 x, you get two solutions: 0 and 1. The check for each solution is shown below.
Check x 5 0. Check x 5 1.
2!x 5 x 2!x 5 x
2!0 0 0 2!1 0 1
0 5 0 21 2 1
4. Which solution is extraneous? Explain.
_______________________________________________________________________
_______________________________________________________________________
HSM12A1MC_1004_NL.indd 302 3/9/11 3:57:12 PM
Cop
yrig
ht ©
by
Pear
son
Educ
atio
n, In
c. o
r its
aff
iliat
es. A
ll Ri
ghts
Res
erve
d.Problem 1
Problem 2
HSM11_A1MC_1004_ T91369
WriteThink
Then I divide each side by 2 to isolate the radical.
Now I square each side.
Next I simplify.
Finally, I multiply each side by 3.3.
= 2√ 𝓵𝓵3.3
= √ 𝓵𝓵3.3
= 𝓵𝓵3.3
= 𝓵𝓵
First I substitute 1 for t in t = 2√ 𝓵𝓵3.3 .
= 2
303 Lesson 10-4
Solving by Isolating the Radical
Got It? What is the solution of !x 2 5 5 22?
5. Circle the first step to solve the equation !x 2 5 5 22. Underline the second step.
Add 5 to both sides. Square both sides. Subtract 22 from both sides.
6. Complete the steps to solve 7. Check your solution by substituting !x 2 5 5 22. for x in the original equation.
5 3 Å 2 5 0 22
x 5 2 5 0 22
0 22
8. The solution I found in Exercise 6 does / does not solve the equation.
Using a Radical Equation
Got It? The time t in seconds it takes for a pendulum of a clock to complete a
full swing is approximated by the equation t 5 2Å<
3.3, where < is the length of the pendulum in feet. If the pendulum of a clock completes a full swing in 1 s, what is the length of the pendulum? Round to the nearest tenth of a foot.
9. Circle what you are trying to find to solve the problem.
length of the pendulum time it takes for 1 swing time it takes for 2 swings
10. Complete the reasoning model below to solve the equation.
11. To the nearest tenth of a foot, the pendulum is ft long.
HSM12A1MC_1004_NL.indd 303 3/9/11 3:57:07 PM
Cop
yrig
ht ©
by
Pear
son
Educ
atio
n, In
c. o
r its
aff
iliat
es. A
ll Ri
ghts
Res
erve
d.
Problem 4
Problem 3
Chapter 10 304
Solving With Radical Expressions on Both Sides
Got It? What is the solution of !7x 2 4 5 !5x 1 10?
12. The equation is solved below. Write a justification for each step.
(!7x 2 4)25 (!5x 1 10)
2
7x 2 4 5 5x 1 10
2x 2 4 5 10
2x 5 14
x 5 7
13. Check your solution.
When you solve an equation by squaring each side, you write a new equation. The new equation may have solutions that do not make the original equation true.
Original Square New Solutions of Equation each side. Equation New Equation
x 5 3 x2 5 32 x2 5 9 3, 23
In the example above, 23 is not a solution of the original equation. It is an extraneous solution.
Identifying Extraneous Solutions
Got It? What is the solution of 2y 5 !y 1 6?
14. Solve the equation.
y2 5 Square each side.
5 0 Subtract y 1 6 from each side.
(y 2 )(y 1 ) 5 0 Factor the quadratic equation.
y 2 5 0 or y 1 5 0 Use the Zero-Product Property.
y 5 or y 5 Solve for y.
15. Check to see whether each result satisfies the original equation.
23 0 Å 1 6 2 0 Å 1 6
23 0 Å 2 0 Å
23 2 2 5
16. The only solution of the equation 2y 5 !y 1 6 is .
HSM12A1MC_1004_NL.indd 304 3/9/11 3:57:04 PM
Cop
yrig
ht ©
by
Pear
son
Educ
atio
n, In
c. o
r its
aff
iliat
es. A
ll Ri
ghts
Res
erve
d.
Lesson Check
Now Iget it!
Need toreview
0 2 4 6 8 10
Math Success
Problem 5
305 Lesson 10-4
• Do you UNDERSTAND?
Vocabulary Which is an extraneous solution of s 5 !s 1 2?
2 0 21 22
20. Solve the equation to find possible 21. Circle the two possible solutions solutions. of the equation.
2 0 21 22
22. Check your solutions to find which is extraneous.
23. The extraneous solution is . The correct answer is C.
Check off the vocabulary words that you understand.
radical equation solution extraneous solution
Rate how well you can solve radical equations.
Identifying Equations With No Solution
Got It? What is the solution of 6 2 !2x 5 10?
17. Solve the equation.
18. Check the solution.
19. The equation does /does not have a solution.
HSM12A1MC_1004_NL.indd 305 3/9/11 3:56:57 PM