Answer:
5 solutions.
x=0
x=0
x=0
x=1
x=-1
Explanation:
To solve this, we need to handle the radical function somehow, an easy approach is to raise to square power the whole equation, this way you can get rid of it.
sqr( x^(3/2) = x^(5/2) )=
x^3 = x^5
Now, lets group terms
x^5-x^3==0
GCF x^3
x^3(x^2-1)=0
The above equation gives us already 5 answer
In order to satisfy the equation there are two choices:
x^3=0 (a)
or (x^2-1)=0 (b)
From (a), we have that x=0, thus given one solution from here.
However, the amount of solutions is not just one, its 3, because the amount of solutions is linked to the exponent of the variable.
the other equation (b)
gives us x =sqrt(1), leading to x= -1 or x=+1
We have two solutions here, again, the exponent of the variable already tells us the amount of solutions,