Analyse the illustrative examples below then try to define an extraneous root. 1. Solve and check: 4 + Vx - 2 = x Checking: If x = 3, 4 + Vx - 2 = 3 4 + V3 -2 = 3 Solution: Extraneous Root 4 + Vx-2 = x Vx -2 x- 4 4 + Vī = 4 + 1 = 3 5# 3 (Vx - 2)2 = (x- 4)² %3D x - 2 = x2 - 8x +16 Checking: If x = 6, 4 + Vx -2 =x 4 + V6 -2 = 6 4 + V4 = 6 4 + 2 = 6 6 = 6 x2 - 9x + 18 0 (x - 3 )(x -6) = 0 This is the only solution. x = 3 x =6 Checking: x - 6 = Vx (x – 6)2 = (Vx )2 x = 9 This is the only solution. x - 6 = Vx 9 -6 = V9 x2 - 12x + 36% = 0 3 = 3 x2 -12x-x+36% 0 x2 - 13x + 36 = 0 Checking: (x - 9)(x- 4) = 0 x = 4 x-6 = Vx 4-6= V4 Extraneous solutions. x = 9, x = 4 -2 # 2 Guide Questions: 1. How would you define an extraneous root based on the illustrative example? 2. What data have you used to define an extraneous root? 3. How is the process of checking related to finalizing your answer to a problem?
Analyse the illustrative examples below then try to define an extraneous root. 1. Solve and check: 4 + Vx - 2 = x Checking: If x = 3, 4 + Vx - 2 = 3 4 + V3 -2 = 3 Solution: Extraneous Root 4 + Vx-2 = x Vx -2 x- 4 4 + Vī = 4 + 1 = 3 5# 3 (Vx - 2)2 = (x- 4)² %3D x - 2 = x2 - 8x +16 Checking: If x = 6, 4 + Vx -2 =x 4 + V6 -2 = 6 4 + V4 = 6 4 + 2 = 6 6 = 6 x2 - 9x + 18 0 (x - 3 )(x -6) = 0 This is the only solution. x = 3 x =6 Checking: x - 6 = Vx (x – 6)2 = (Vx )2 x = 9 This is the only solution. x - 6 = Vx 9 -6 = V9 x2 - 12x + 36% = 0 3 = 3 x2 -12x-x+36% 0 x2 - 13x + 36 = 0 Checking: (x - 9)(x- 4) = 0 x = 4 x-6 = Vx 4-6= V4 Extraneous solutions. x = 9, x = 4 -2 # 2 Guide Questions: 1. How would you define an extraneous root based on the illustrative example? 2. What data have you used to define an extraneous root? 3. How is the process of checking related to finalizing your answer to a problem?
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
Related questions
Question
Transcribed Image Text:Analyse the illustrative examples below then try to define an extraneous root.
Checking:
If x = 3,
4 + Vx -2 = 3
4 + V3 -2 = 3
4 + Vĩ = 3
4 + 1 = 3
1. Solve and check:
4 + Vx-2 = x
Solution:
%3|
Extraneous Root
4 + Vx - 2 = x
%3D
%3D
Vx - 2 %3D х-4
5# 3
(Vx – 2)? = (x – 4)²
%3D
x - 2 = x2 - 8x +16
Checking:
If x = 6,
x2 - 9x + 18 0
%3D
%3D
4 + Vx -2 = x
(x – 3 )(x -6) = 0
This is the only solution.
4 + v6 - 2 = 6
4 + V4 = 6
x = 3 x = 6
4 + 2 = 6
6 = 6
Checking:
x-6 = Vx
%3D
X = 9
x - 6 = Vx
9 – 6 = V9
(x – 6)2 = (Vx)2
This is the only solution.
%3D
%3D
x2 -12x + 36 = 0
%3D
3 = 3
x2 – 12x - x +36 = 0inw
%3D
x2 - 13x +36 = 0
Checking:
(x - 9)(x- 4) = 0
x = 4
x - 6 = Vx
4 - 6 = V4
%3D
Extraneous solutions.
x = 9,
X = 4
%3D
-2 # 2
Guide Questions:
1. How would you define an extraneous root based on the illustrative example?
2. What data have you used to define an extraneous root?
3. How is the process of checking related to finalizing your answer to a problem?
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