confusing algebra problem please help me im stuck =(?

Question

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Answer 1

The solution drew c found is correct, but there are no choices given that are correct:
ex·tra·ne·ous
Pronunciation: \ek-ˈstrā-nē-əs\
Function: adjective
Etymology: Latin extraneus — more at strange
Date: 1638
1: existing on or coming from the outside
2 a: not forming an essential or vital part b: having no relevance
3: being a number obtained in solving an equation that is not a solution of the equation

The one solution is unique, not extraneous.

±√((-29/2)^2 - 28) - 1 =? - 29/2
±√(841/4 - 28) - 1 =? - 29/2
±√((841 - 112)/4) - 1 =? - 29/2
±(1/2)√(729) - 1 =? - 29/2
±(27/2) - 1 =? - 29/2
- 29/2 = - 29/2

While √ without a sign is usually assumed to imply the positive root, there are cases, like this one, where the negative root is the functional root.

Answer 2

solving √(x²-28) - 1 = x
√(x²-28) = x + 1
x² - 28 = x² + 2x + 1
-29 = 2x
-29/2 = x
looks like 1 real solution, but
√(x²-28) cannot be negative, and x+1 = -27/2 is,
so there are NO real solutions.
When you square both sides, extraneous solutions creep in, so any solutions you get must be checked back into the original equation.

Answer 3

==== Ignore this answer, it is incorrect... see the other answers below for better understanding. ====

sqrt(x^2-28) - 1 = x ___ First move the 1 to the right side
sqrt(x^2-28) = x + 1 ___Next square both sides
x^2-28 = (x+1)^2 ____ Next simplify the right side
x^2 - 28 = x^2 + 2x + 1 _____Next cancel out the x^2
-28 = 2x + 1

At this point it becomes evident that there is one solution (One extraneous solution)... just for kicks, this solution is

x = -29/2

Answer 4

in case you prefer to get A*(x1,x2,x3)^T = (x2,x3,x1)^T you may get A as follows: Row a million of A is the 2d row of the identity matrix, row 2 is the third row and row 3 is the 1st row. So A is the matrix 0 a million 0 0 0 a million a million 0 0 examine via multiplying A via (x1, x2, x3)^T. observe: The T right here ability transpose, i.e., write this as a column vector.