lesson 3 menu five-minute check (over lesson 10-2) main ideas and vocabulary targeted teks example...
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![Page 1: Lesson 3 Menu Five-Minute Check (over Lesson 10-2) Main Ideas and Vocabulary Targeted TEKS Example 1: Variable in Radical Example 2: Radical Equation with](https://reader035.vdocuments.net/reader035/viewer/2022062422/56649f115503460f94c235e9/html5/thumbnails/1.jpg)
Five-Minute Check (over Lesson 10-2)
Main Ideas and Vocabulary
Targeted TEKS
Example 1: Variable in Radical
Example 2: Radical Equation with an Expression
Example 3: Variable on Each Side
![Page 2: Lesson 3 Menu Five-Minute Check (over Lesson 10-2) Main Ideas and Vocabulary Targeted TEKS Example 1: Variable in Radical Example 2: Radical Equation with](https://reader035.vdocuments.net/reader035/viewer/2022062422/56649f115503460f94c235e9/html5/thumbnails/2.jpg)
• radical equation
• extraneous solution
• Solve radical equations.
• Solve radical equations with extraneous solutions.
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FREE-FALL HEIGHT An object is dropped from an unknown height and reaches the ground in 5 seconds. From what height is it dropped?
Variable in Radical
Original equation
Replace t with 5.
Multiply each side by 4.
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Square each side.
400 = h Simplify.
Check by substituting 400 for h in the original equation.
Answer: 400 ft
Variable in Radical
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A. A
B. B
C. C
D. D A B C D
0% 0%0%0%
A. 28 ft
B. 11 ft
C. 49 ft
D. 784 ft
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Radical Equation with an Expression
x = 52 Add 3 to each side.
Answer: The solution is 52.
Original equation
Subtract 8 from each side.
Square each side.
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Variable on Each Side
2 – y = y2 Simplify.
0 = y2 + y – 2Subtract 2 and add y to each side.
0 = (y + 2)(y – 1)Factor.
y + 2 = 0 or y – 1 = 0Zero Product Property
y = –2 y = 1Solve.
Original equation
Square each side.
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Variable on Each Side
Check
Answer: Since –2 does not satisfy the original equation, 1 is the only solution.
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