1. What positive number is less than twice its square by the cube-root of two?

Answer 1

No such integer number exists. What you are asking is

n = 2 n^2 - cuberoot2 or

2n^2 - n - cuberoot 2 = 0

Use the quadratic equation formula to calculate the roots n1 and n2.

It is not an integer as you can see.

Answer 2

x = positive number

x = 2x^2 - 2^(1/3)

0 = 2x^2 - x - 2^(1/3)

x = [ -(-1) +or- ( (-1)^2 - 4(2)(2)^(1/3) )^(1/2) ] / [ 2(2) ]

x = [ 1 +or- ( 1 - 2^(10/3) )^(1/2) ] / 4

x = 1 + 0.7533i or x = 1 - 0.7533i approximatly

Complex numbers are neither positive nor negative. No positive number is less than twice its square by the cube-root of two.

Answer: This equation has no solution.

Answer 3

you got me there
thank for the two