To solve the problem of finding how many different values of [tex]\( x \)[/tex] make [tex]\(\sqrt{\frac{48}{x}}\)[/tex] a whole number, we need to consider the expression [tex]\(\sqrt{\frac{48}{x}}\)[/tex].
When [tex]\(\sqrt{\frac{48}{x}}\)[/tex] is a whole number, let's denote this whole number by [tex]\( k \)[/tex]. Therefore, we can set up the equation:
[tex]\[
\sqrt{\frac{48}{x}} = k
\][/tex]
Squaring both sides to eliminate the square root, we have:
[tex]\[
\frac{48}{x} = k^2
\][/tex]
Rearranging the equation to solve for [tex]\( x \)[/tex], we get:
[tex]\[
x = \frac{48}{k^2}
\][/tex]
Here, [tex]\( x \)[/tex] has to be a positive integer. Thus, [tex]\(\frac{48}{k^2}\)[/tex] must also be a positive integer, meaning [tex]\( k^2 \)[/tex] must be a divisor of 48.
To find suitable [tex]\( k \)[/tex], we first determine the divisors of 48. The prime factorization of 48 is:
[tex]\[
48 = 2^4 \times 3
\][/tex]
The divisors of 48 are obtained by considering all combinations of its prime factors:
[tex]\[
1, 2, 3, 4, 6, 8, 12, 16, 24, 48
\][/tex]
Since [tex]\( k^2 \)[/tex] must be a divisor of 48, we check the divisors of 48 that are perfect squares. The perfect square divisors of 48 are:
[tex]\[
1, 4, 16
\][/tex]
Thus, the corresponding [tex]\( k \)[/tex] values (for which [tex]\( k^2 \)[/tex] is a divisor of 48) are:
[tex]\[
k = 1, 2, 4
\][/tex]
Now, substituting these [tex]\( k \)[/tex] values back into our equation:
[tex]\[
x = \frac{48}{k^2}
\][/tex]
- For [tex]\( k = 1 \)[/tex]:
[tex]\[
x = \frac{48}{1^2} = \frac{48}{1} = 48
\][/tex]
- For [tex]\( k = 2 \)[/tex]:
[tex]\[
x = \frac{48}{2^2} = \frac{48}{4} = 12
\][/tex]
- For [tex]\( k = 4 \)[/tex]:
[tex]\[
x = \frac{48}{4^2} = \frac{48}{16} = 3
\][/tex]
Therefore, the different values of [tex]\( x \)[/tex] that satisfy the given condition are [tex]\( x = 3, 12, \)[/tex] and [tex]\( 48 \)[/tex].
Hence, there are [tex]\(\boxed{3}\)[/tex] different values of [tex]\( x \)[/tex] for which [tex]\(\sqrt{\frac{48}{x}}\)[/tex] is a whole number.
