For how many real values of x is sqrt(63- sqrt(x) an integer?

To solve this problem, we need to determine the values of x that make the expression sqrt(63 - sqrt(x)) an integer.

Let's break down the problem into steps:

Step 1: Simplify the expression inside the square root:
63 - sqrt(x)

Step 2: Since sqrt(63 - sqrt(x)) should be an integer, the expression 63 - sqrt(x) must also be a perfect square. Therefore,
63 - sqrt(x) = y^2, where y represents the integer value.

Step 3: Re-arrange the equation to isolate sqrt(x):
sqrt(x) = 63 - y^2

Step 4: Square both sides of the equation to eliminate the square root:
x = (63 - y^2)^2

Step 5: Expand the equation:
x = 63^2 - 2y^2(63) + y^4

Now, we need to count the number of possible real values of x that satisfy the given conditions.

To find these values, we need to determine the number of perfect squares that are less than or equal to 63^2. Let's consider each perfect square less than or equal to 63^2:

1^2 = 1
2^2 = 4
3^2 = 9
4^2 = 16
...
8^2 = 64 (greater than 63^2)

From this, we can see that there are 8 perfect squares less than or equal to 63^2.

Therefore, there are 8 possible real values of x that make sqrt(63 - sqrt(x)) an integer.

To find the number of real values of x for which sqrt(63 - sqrt(x)) is an integer, we can follow these steps:

Step 1: Start by simplifying the expression sqrt(63 - sqrt(x)). Notice that the term sqrt(63 - sqrt(x)) will only be an integer if the quantity 63 - sqrt(x) itself is a perfect square.

Step 2: Set 63 - sqrt(x) equal to a perfect square, let's call it n^2.

63 - sqrt(x) = n^2

Step 3: Solve for sqrt(x) by isolating it on one side of the equation.

sqrt(x) = 63 - n^2

Step 4: Square both sides of the equation to get rid of the square root.

x = (63 - n^2)^2

Step 5: Determine how many different values of n can satisfy the equation x = (63 - n^2)^2. Since n can take any integer value, we need to find all possible perfect squares n^2 that are less than or equal to 63.

The perfect squares less than or equal to 63 are: 0, 1, 4, 9, 16, 25, 36, 49, and 64 (which is greater than 63). So there are 8 possible values of n.

Step 6: Substitute each value of n into the equation x = (63 - n^2)^2 to find the corresponding values of x.

For n = 0, x = (63 - 0^2)^2 = 63^2 = 3969
For n = 1, x = (63 - 1^2)^2 = 62^2 = 3844
For n = 2, x = (63 - 2^2)^2 = 59^2 = 3481
For n = 3, x = (63 - 3^2)^2 = 54^2 = 2916
For n = 4, x = (63 - 4^2)^2 = 47^2 = 2209
For n = 5, x = (63 - 5^2)^2 = 38^2 = 1444
For n = 6, x = (63 - 6^2)^2 = 27^2 = 729
For n = 7, x = (63 - 7^2)^2 = 14^2 = 196

So, there are 8 real values of x for which sqrt(63 - sqrt(x)) is an integer. The values of x are 3969, 3844, 3481, 2916, 2209, 1444, 729, and 196.

TWICE ONCE

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