Question #26706

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Answer 1 · 2017-11-12T01:21:39

change back into radical form, and solve.

$sqrt(x^2-1) = 0$

we can square both sides because of the square root property

$(sqrt(x^2-1))^2 = (0)^2$

$x^2-1=0$

$x^2-1$ is a difference of squares

$(x-1)(x+1) = 0$

we take each factor and equal it to $0$

$(x-1 )= 0$
$x = 1$

$(x+1) = 0$
$x = -1$

so our domain is $x<= -1$ and $x>=1$ which is the set of $RR$

to be clear $0$ is not included because Domain corresponds to the X values where y is $0$

Side note if 0 was allowed, $sqrt(0^2-1)$ will be $sqrt(-1)$ which is an imaginary value, and not in the set of $RR$

set builder ${x in RR | x<=-1 or x>=1}$