1.02C - Square Root Equations | High School

University:
High School
Course:
High School Physics
Academic year:

2024
Views:

439

Pages:

16
Author:

1 1

SQUARE ROOT FUNCTIONS 1.02C – SOLVING EQUATIONS • Follow Order of Operations Ex 1: 𝑓 𝑥 = 1 4 3(𝑥 − 7) + 2 Find: 𝑓 19 7 𝑓 19 = 2 Ex 2: 𝑓 𝑥 = −2 −𝑥 + 1 − 3 Find: 𝑓 −161 𝑓 −161 = −18 2 − 3 Finding 𝑦 given 𝑥 • Steps: 1. Identify the Restrictions (if not given) 2. 3. 4. 5. Isolate the radical 𝑏 𝑥−ℎ Square each side to eliminate radical symbol Solve Check Finding 𝑥 given 𝑦 Find the zero of the function 𝑓 defined by 1. State the restrictions --SET TO ZERO-2. Isolate the radical: 3. Square each side to eliminate the radical 4. Solve 5. Check Ex 1 𝒇 𝒙 =𝟐 𝒙−𝟑−𝟒 Restrictions: 𝑥 − 3 ≥ 0 𝑥≥3 2 𝑥−3−4=0 2 𝑥−3=4 𝑥−3=2 ( 𝑥 − 3)2 = (2)2 𝑥−3=4 𝑥=7 7 ≥ 3 and 7 verifies 2 𝑥 − 3 − 4 = 0 ∴ 7 𝑖𝑠 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 Find the zero of the function 𝑓 defined by 1. State the restrictions --SET TO ZERO-2. Isolate the radical: 3. Square each side to eliminate the radical 4. Solve 5. Check Ex 2 𝒇 𝒙 = −𝟐 − 𝒙 − 𝟒 + 𝟐 Restrictions: −(𝑥 − 4) ≥ 0 𝑥≤4 −2 − 𝑥 − 4 + 2 = 0 −2 −(𝑥 − 4) = −2 −(𝑥 − 4) = 1 ( −(𝑥 − 4))2 = (1)2 −(𝑥 − 4) = 1 𝑥=3 3 ≤ 4 and 3 verifies −2 − 𝑥 − 4 + 2 = 0 ∴ 3 𝑖𝑠 𝑡ℎ𝑒 𝑧𝑒𝑟𝑜 𝑜𝑓 𝑡ℎ𝑒 𝑓𝑢𝑛𝑐𝑡𝑖𝑜𝑛 Find the zero of function 𝑓 defined by R: 𝑥 ≥ −3 𝒇 𝒙 =𝟐 𝒙+𝟑+𝟐 2 𝑥+3+2=0 2 𝑥 + 3 = −2 𝑥 + 3 = −1 2 Ex 3 𝑥 + 3 = −1 2 𝑥+3=1 𝑥 = −2 TEST: 2 −2 + 3 + 2 = 0 4≠0 IMPOSSIBLE!! 𝑎𝑛𝑦𝑡ℎ𝑖𝑛𝑔 ≠ 𝑛𝑒𝑔𝑎𝑡𝑖𝑣𝑒‼ REMEMBER now • Find 𝑓 𝑥 = 𝑔 𝑥 , where 𝑓 𝑥 =2 − 𝑥−3 −4 𝑔 𝑥 = −8 Restrictions: 𝑥≤3 𝑥= ∅ Ex 5 Solve the equation: 2 𝑥 + 4 − 1 = 𝑥 2 𝑥+4−1=𝑥 2 𝑥+4=𝑥+1 𝑥+1 𝑥+4= 2 Restrictions: (1) 𝑥 + 4 ≥ 0 ⇒ 𝑥 ≥ −4 (2) 𝑥 + 1 ≥ 0 ⇒ 𝑥 ≥ −1 𝑇ℎ𝑒𝑟𝑒𝑓𝑜𝑟𝑒 𝑡ℎ𝑒 𝑒𝑞𝑢𝑎𝑡𝑖𝑜𝑛 𝑟𝑒𝑞𝑢𝑖𝑟𝑒𝑠 𝑥 ≥ −1 Ex 6 𝑥+1 𝑥+4= 2 2 𝑥+1 2 ( 𝑥 + 4) = 2 𝑥 2 + 2𝑥 + 1 𝑥+4= 4 2 4𝑥 + 16 = 𝑥 + 2𝑥 + 1 𝑥 2 − 2𝑥 − 15 = 0 Ex 6 𝑥 2 − 2𝑥 − 15 = 0 ------------------------FACTOR----------------------𝑥−5 𝑥+3 =0 𝑥−5=0 or 𝑥+3=0 𝑥=5 or 𝑥 = −3 −3 ≤ −1 rejected! ∴𝑥=5 Ex 6 𝑉: 2 (5) + 4 − 1 = (5) 5=5 Solve 𝟐 𝟐𝒙 + 𝟏 = 𝟑 𝒙 4(2𝑥 + 1) = 9𝑥 8𝑥 + 4 = 9𝑥 4=𝑥 Restrictions: (1) 𝑥 ≥ 0 1 (2) 2𝑥 + 1 ≥ 0 → 𝑥 ≥ − 2 Ex 7 𝑉: 2 2 4 + 1 = 3 4 6=6 Solve −𝟐 − 𝒙 − 𝟓 + 𝟒 = 𝟐 𝒙 + 𝟓 − 𝟒 −2 −(𝑥 − 5) + 8 = 2 (𝑥 + 5) − −(𝑥 − 5) + 4 = (𝑥 + 5) Square the sides to get rid of at least ONE radical (− − 𝑥 − 5 + 4)(− − 𝑥 − 5 + 4) = 𝑥 + 5 −𝑥 + 5 − 8 − 𝑥 − 5 + 16 = 𝑥 + 5 −8 − 𝑥 − 5 = 2𝑥 − 16 Ex 8 −8 − 𝑥 − 5 = 2𝑥 − 16 Square again! 64 −𝑥 + 5 = 4𝑥 2 − 64𝑥 + 256 −64𝑥 + 320 = 4𝑥 2 − 64𝑥 + 256 0 = 4𝑥 2 − 64 0 = (𝑥 2 − 16) 0 = (𝑥 + 4)(𝑥 − 4) 𝑥 = {−4, 4} Ex 8 Check! −𝟐 − 𝒙 − 𝟓 + 𝟒 = 𝟐 𝒙 + 𝟓 − 𝟒 Restrictions: − 𝒙 − 𝟓 ≥ 0 and 𝒙 + 𝟓 ≥ 𝟎 𝒙 ≤ 5 and 𝒙 ≥ −𝟓 Validations: 𝒙 = −𝟒 ∶ −𝟐 −(−𝟒 − 𝟓) + 𝟒 = 𝟐 −𝟒 + 𝟓 − 𝟒 −𝟐 = −𝟐 𝒙=𝟒 ∶ −𝟐 −(𝟒 − 𝟓) + 𝟒 = 𝟐 𝟒 + 𝟓 − 𝟒 𝟐=𝟐 ∴ 𝑥 = {−4, 4} are the solutions. •Visions 1 P39: #4acdh P42: #13ad P43: #16 Homework: