Question

2x} }{\sqrt{} x-1}" alt="\frac{\sqrt{2x} }{\sqrt{} x-1}" align="absmiddle" class="latex-formula">
For which values of x does each expression make sense?

Answer 1

Answer:

The expression is unclear so:

If the denominator is $\sqrt{x}-1$ then your set of existence is given by

$\left \{ {{x\ge0} \atop {x\ne1}} \right.$

since you want the quantity inside both square roots to be positive (and it happens for both to be simply $x\ge0$ and you don't want a zero at the denominator so you have to rule out 1.

If the square root includes the whole denominator $\sqrt{x-1}$ the condition becomes $\left \{ {{x\ge0} \atop {x>1}} \right.$

where you don't include the extreme in the second condition since, again, you don't want to divide by 0. The answer in this case is simply $x>1$ since values between 0 and 1 will give a negative root which is not a real number.