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Find all solutions of the equation |x^2 - 14x + 29| = 4. Discuss whether or not your solution generates extraneous solutions.
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You need to solve actually two quadratic equations
1) x^2 - 14x + 29 = 4, (1) and
2) x^2 - 14x + 29 = -4 (2)
independently, and then to take the union of their roots as a set of solutions to the absolute value equation.
So, start from eq(1).
x^2 - 14x + 29 = 4 ====> x^2 - 14x + (29-4) = 0 ====>
x^2 - 14x + 25 = 0 ====> = = = = .
Now work on eqn(2)
x^2 - 14x + 29 = -4 ====> x^2 - 14x + (29 +4) = 0 ====>
x^2 - 14x + 33 = 0 ====> = = = .
So, the solutions to your original equation are these four real numbers:
, , and .
Plot y = and y = 4.
These four solutions are x-coordinates of intersection points.
My solution did not produce extraneous solutions.
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On solving absolute value equations see the lessons
- Absolute Value equations
- HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 1
- HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 2
- HOW TO solve equations containing Linear Terms under the Absolute Value sign. Lesson 3
- HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 1
- HOW TO solve equations containing Quadratic Terms under the Absolute Value sign. Lesson 2
in this site.
Also, you have this free of charge online textbook in ALGEBRA-I in this site
- ALGEBRA-I - YOUR ONLINE TEXTBOOK.
The referred lessons are the part of this online textbook under the topic
"Solving Absolute values equations".
Save the link to this online textbook together with its description
Free of charge online textbook in ALGEBRA-I
https://www.algebra.com/algebra/homework/quadratic/lessons/ALGEBRA-I-YOUR-ONLINE-TEXTBOOK.lesson
to your archive and use it when it is needed.