Let the integers be x-2, x, x+2.
Given,
x -2 + x + x + 2 = 69
3x = 69
x = 23
Hence, the integers are 21, 23 and 25.
Set the whole expression = to 0 and solve for x.
3x^(5/3) - 4x^(7/3) = 0. Factor out x^(5/3): x^(5/3) [3 - 4x^(2/3)] = 0
Then either x^(5/3) = 0, or 3 - 4x^(2/3) = 0.
In the latter case, 4x^(2/3) = 3.
To solve this: mult. both sides by x^(-2/3). Then we have
4x^(2/3)x^(-2/3) = 3x^(-2/3), or 4 = 3x^(-2/3). It'd be easier to work with this if we rewrote it as
4 3
--- = --------------------
1 x^(+2/3)
Then
4
--- = x^(-2/3). Then, x^(2/3) = (3/4), and x = (3/4)^(3/2). According to my 3 calculator, that comes out to x = 0.65 (approx.)
Check this result! subst. 0.65 for x in the given equation. Is the equation then true?
My method here was a bit roundabout, and longer than it should have been. Can you think of a more elegant (and shorter) solution?
1220/8= 152.5
So 8 can go into 1220 152.5 times
