\frac{√(2x) }{\sqrt{} x-1} For which values of x does each expression make sense?

Question

`\frac{√(2x) }{\sqrt{} x-1}`

For which values of x does each expression make sense?

Answer 1

The expression (√2x) / (√x - 1) is valid for x > 1, to ensure both the radicand and denominator are well-defined and non-negative.

Step-by-step explanation:

The student's question involves determining for which values of x the expression (√2x) / (√x - 1) makes sense. The expression is valid for real numbers as long as the conditions that the denominator and radicand must be non-negative are met. The denominator (√x - 1) suggests that x must be greater than 1, which ensures that the square root is defined and that the denominator is not zero, which would make the expression undefined. Moreover, the numerator (√2x) implies that x must be non-negative to ensure the radicand is non-negative, and thus, x should be greater than or equal to 0.

However, since the expression must satisfy both conditions simultaneously, the values of x that make the expression valid should be greater than 1. To summarize, the expression makes sense for x > 1.

Answer 2

The expression is unclear so:

If the denominator is
`√(x)-1` then your set of existence is given by

`\left \{ {{x\ge0} \atop {x\\e1}} \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
`√(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.