Question

How do you solve `(8x)^(1/2)+6=0`?

Answer 1

`x=9/2`

`x=4.5`

Explanation:

`(8x)^(1/2)+6=0`

Get rid of 6 from left side
For that subtract 6 on both sides

`(8x)^(1/2)=-6`

Squaring on both sides

`8x=36`

`x=36/8`

`x=9/2`

`x=4.5`

Answer 2

There are no values of `x` which satisfy this equation.

Explanation:

`(8x)^(1/2)+6=0`

Subtract `6` from both sides to get:

`(8x)^(1/2) = -6`

Square both sides, noting that squaring may introduce spurious solutions:

`8x = 36`

Divide both sides by `8` to get:

`x = 36/8 = 9/2`

Check:

`(8x)^(1/2)+6 = (8*9/2)^(1/2)+6 = 36^(1/2)+6 = 6+6 = 12`

So this `x` is not a solution of the original equation.

The problem is that while `36` has two square roots (viz `+-6`), `36^(1/2) = sqrt(36) = 6` denotes the principal, positive square root.

So the original equation has no solutions (Real or Complex).