Need help with this math question!! Please!!!?

x^(1 / 2) + 2[(x / 1)^(-1 / 2)] = 15[(x / 1)^(-3 / 2)]

x^(1 / 2) + 2[(1 / x)^(1 / 2)] = 15[(1 / x)^(3 / 2)]

x^(1 / 2) + 2[1 / x^(1 / 2)] = 15[1 / x^(3 / 2)]

Note: x^(3 / 2) = √(x^3) = √(x * x * x) = √(x^2) * √x = x√x,
as half power equals square root:

√x + (2 / √x) = 15 / (x√x)
Multiply everything by x√x to linearize this, and clear the denominator. By linearize, I just mean put everything on one line.

(√x)(x√x) + (2 / √x)(x√x) = (15 / x√x)(x√x)
Note: (√x)(√x) = √(x * x) = √(x^2) = x

x^2 + 2x = 15
Set equal to zero by subtracting 15 to both sides:

x^2 + 2x - 15 = 0
Factor, as -3 * 5 = -15, and -3 + 5 = 2. Boom!

(x - 3)(x + 5) = 0
Set each quantity equal to zero:

x - 3 = 0, so x = 3

x + 5 = 0, so x = -5
But, we have to dump x = -5 because (-5)^(1 / 2) or √(-5) gives us an imaginary number.

Solution: x = 3
-------
This is hard because you have rational powers. However, with some manipulation, you can make it a quadratic, which I'm sure you know how to solve.

multiply by x^(3/2), make a quadratic
x*2 + 2x - 15 = 0
(x - 3)(x + 5)=0
x=3 or x=-5 reject -5 because of sqrt(x)

the answer is 3

x^1/2 + 2x^-1/2 = 15x^-3/2 ( multiply by x ^3/2 ) , we get,
x^1/2+3/2 + 2x-1/2+ 3/2 = 15x^-3/2+ 3/2 ,
x2 + 2x = 15x^0 ,
x2 +2x = 15 ( since x^0 = 1 ) ,
so x2 +2x -15 + 0 ,
which is; x2 +5x - 3x -15 =0 ( factorization : 5 *3 = 15 , and + 5 -3 =2 ) ,
that is, x ( x+5 ) -3( x+ 5 ) =0,
( x+5 ) ( x-3 ) = 0 , so x = -5 or +3 , now substitute x = -5 in the equation x2 +2x -15 , we get ( -5 )2 +2 ( -5 ) -15 = 25 -10 -15 =0 , which we get on the right hand side of the equation ,
similarly substitute x = +3 in the equation , and we get value as zero ( 0 ) , which is complying as right hand side of the equation is also zero , ths x = -5 0r + 3 is complying and therefore answer is correct

Put u = 1/x

x^(1/2) + 2x^(−1/2) = 15x^(-3/2)
[x^(1/2) + 2x^(−1/2)] • x^(-1/2) = 15x^(-3/2) • x^(-1/2)
1 + 2x^(-1) = 15x^(-2)
1 + 2(1/x) = 15(1/x)^2
1 + 2u = 15u^2
15u^2 - 2u - 1 = 0
(3u - 1)(2u + 1) = 0
u = 1/3 or u = -1/2
x = 3 or x = -2 (rejected)

[x = -2 is rejected because (-2)^(-1/2) is NOT real.]

x^1/2 + 2x^−1/2 = 15x^−3/2

multiple by x^(3/2), add the exponents

x^2 +2x^1 = 15x^0
x^2 +2x -15 = 0
3 & 5 are factors, middle is positive so +5 -3
(x+5)(x-3) = 0

x = -5, x=3 are both answers.

multiply both sides by x^(3/2) to generate a quadratic , solve it