Question

Question #cabf2

Answer 1

I found two real solutions:

Explanation:

I would try rearranging it as:
`x^2=4sqrt(x)`
square both sides:
`x^4=16x`
`x^4-16x=0`
`x(x^3-16)=0`
so we have:
`x=0`
and
`x^3-16=0`
`x^3=16`

solutions:
`x_1=0`
`x_2=root(3)(16)`

`x_3` and `x_4` should be two imaginary solutions considering that the graph of `x^3-16` has only one real solution (`x` intercept):
graph{x^3-16 [-105.4, 105.4, -52.7, 52.8]}

Answer 2

`x=2.52`

Explanation:

`x^2-4sqrtx=0`

`x^2=4sqrtx`

`x^2/sqrtx=x^(3/2)=4`

`x^(3/2)=(sqrtx)^3=4`

`sqrtx=root(3)4`

`x=(root(3)4)^2=2.52`