For how many real values of x is sqrt(120 - sqrt(x^2)) an integer?
First note that sqrt(x2) is simply x. But wait, it's not simply x, it is +x. Aha.
If you want to consider only the positive values of sqrt(x2), then there are these:
sqrt (120 – 20) = 10
sqrt (120 – 39) = 9
sqrt (120 – 56) = 8
sqrt (120 – 71) = 7
sqrt (120 – 84) = 6
sqrt (120 – 95) = 5
sqrt (120 – 104) = 4
sqrt (120 – 111) = 3
sqrt (120 – 116) = 2
sqrt (120 – 119) = 1 Looks like 10 of them. If you count them this way.
But if you want to take into account the negative root of x2 then there is no finite number of values.
For example, sqrt (120 – (–1)) = sqrt (121) = 11
sqrt (120 – (–24)) = sqrt (144) = 12
sqrt (120 – (–49)) = sqrt (169) = 13
As you can see, this will go on forever, so there is no numerical total.
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