For how many non-negative real values of x is \(\sqrt{144-\sqrt[3]{x}}\) an integer?
The expression: Sqrt(144 - x^(1/3)) is an integer for the following values of x:
x=0, 12,167, 85,184, 250,047, 512,000, 857,375, 1,259,712, 1,685,159, 2,097,152, 2,460,375, 2,744,000, 2,863,288, 2,924,207 = 13 real values of x.
To have an integer result.....the quantity under the root must be a perfect square
So...we will have 13 values
To see this.....when x = 0 the result will be 12
When x = (144)^3 the result will be 0
So.....we can generate all the integers from 0 - 12 by taking the square root of the quantity under the radical .....so.....13 values of x
Here's one more way to see this
Let us suppose that the result of taking the square root is some integer > 12
Then....when we square both sides we have that
144 - (x)^(1/3) = (minimum of 169)
Subtract 144 from both sides
-(x)^(1/3) = (minimum of 25)
Call the right side some positive integer, P
- (x)^(1/3) = P
(x)^(1/3) = -P
(x)^(1/3) = (-1)P cube both sides
x = (-1)^3 P^3
x = - P^3
So.....x is negative.....but we require that x is positive....so.....the result of evaluating the original expression cannot be an integer > 12
Now suppose that is some integer between 0 - 12 inclusive
So.....when we square both sides......the max = 144 and the min = 0
If the max is 144 then
144 - (x)^(1/3) = 144
-(x)^(1/3) = 0
x^(1/3) = 0
And x = 0 so x is non-negative
If the min is 0
144 - (x)^(1/3) = 0
-(x)^(1/3) = - 144
x^(1/3) = 144
x = 144^3 and x is also non-negative
So x will be positive whenever the original quantity under the root is a perfect square and the right side is an integer from 0 to12 inclusive