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Algebra
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Find the number of real solutions for x^2-1=0
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First, factor. This is a difference of squares, so the factorization is (x+1)(x-1) = 0. Do you see what to do next?
x^-1?
Do you mean x^2 - 1?
Oh.
You mean x^-1, or 1/x.
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yea,
x^2-1
You want to factor that, right? So because it's a difference of squares, you factor it like (x+1)(x-1) = 0. Does that make sense?
The square root of any number in +- , two roots and if your taking the square root of a positive real number the result will be a real number
a real number or 2
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It will be 2 because the equation is quadratic and this equation happens to have 2 solutions.
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