How to deal with two solutions of x that are the same?

[iwashere]

I’ve been working on this question related to reducible quadratic equations and have ended up with 4 solutions i.e. x=1, x=1, x=-1, x=-1
now, I am aware that general rule of sets does not allow repetition of numbers within the set. So, how do I construct the solution set for my question based on these results?

for further refrence, following is the reduced form of the equation I was dealing with.
x^4 - 2x^2 + 1 = 0
this is how I went about solving it:
(x^2)^2 - 2(x^2)(1) + (1^2) = 0 i.e. a^2 + 2(a)(b) + (b^2) = 0
(x^2 - 1)^2 = 0
(x^2-1)(x^2-1) = 0
(x+1)(x-1)(x+1)(x-1) = 0 i.e. a^2 - b^2 = (a+b)(a-b)
therefore, x=1, x=1, x=-1, x=-1

how to construct the solution set now?

 

[Deleted member 4993]

I’ve been working on this question related to reducible quadratic equations and have ended up with 4 solutions i.e. x=1, x=1, x=-1, x=-1
now, I am aware that general rule of sets does not allow repetition of numbers within the set. So, how do I construct the solution set for my question based on these results?

for further refrence, following is the reduced form of the equation I was dealing with.
x^4 - 2x^2 + 1 = 0
this is how I went about solving it:
(x^2)^2 - 2(x^2)(1) + (1^2) = 0 i.e. a^2 + 2(a)(b) + (b^2) = 0
(x^2 - 1)^2 = 0
(x^2-1)(x^2-1) = 0
(x+1)(x-1)(x+1)(x-1) = 0 i.e. a^2 - b^2 = (a+b)(a-b)
therefore, x=1, x=1, x=-1, x=-1

how to construct the solution set now?

What was EXACTLY asked of you?

 

[iwashere]

I ask regarding the constriction of solution set with the answers
x=1 x=1 x=-1 x=-1
can values be repeated in a solution set

 

[Deleted member 4993]

I ask regarding the constriction of solution set with the answers
x=1 x=1 x=-1 x=-1
can values be repeated in a solution set

Is this question in the assignment or question posed by yourself?

 

[Steven G]

The solution is a set and in a set it is customary not to repeat anything. The solution is x = +1 or -1.
The way you write this solution depends on what you learned.

{ x | x=1 or x=-1}, {1, -1} or x = 1 or -1 are all ways of writing the solution.

 

[Dr.Peterson]

I’ve been working on this question related to reducible quadratic equations and have ended up with 4 solutions i.e. x=1, x=1, x=-1, x=-1
now, I am aware that general rule of sets does not allow repetition of numbers within the set. So, how do I construct the solution set for my question based on these results?

for further refrence, following is the reduced form of the equation I was dealing with.
x^4 - 2x^2 + 1 = 0
this is how I went about solving it:
(x^2)^2 - 2(x^2)(1) + (1^2) = 0 i.e. a^2 + 2(a)(b) + (b^2) = 0
(x^2 - 1)^2 = 0
(x^2-1)(x^2-1) = 0
(x+1)(x-1)(x+1)(x-1) = 0 i.e. a^2 - b^2 = (a+b)(a-b)
therefore, x=1, x=1, x=-1, x=-1

how to construct the solution set now?

Clearly, as you indicated, the solution set is {-1. 1}. You don't repeat numbers in a set.

You could also be asked to list solutions and their multiplicity; that is a different question, but important. In that case, you could say "the solutions are 1 with multiplicity 2, and -1 with multiplicity 2".

What you say depends on the wording of the problem, which is why you have been asked twice for the actual wording of the problem.