+0

# help

0
105
3
+1195

For how many non-negative real values of x is $$\sqrt{144-\sqrt[3]{x}}$$ an integer?

Oct 15, 2019

#1
+1

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.

Oct 15, 2019
#2
+106536
+1

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

Oct 15, 2019
edited by CPhill  Oct 15, 2019
#3
+106536
+1

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

Oct 15, 2019