Question

How do you solve `sqrt(2x - 8) = 4 -x`?

Answer 1

`x=4`

Explanation:

Start with a condition for existing of a square root on the left side of an equation:
`2x-8 >= 0` or `x >=4`
AND
a condition for the right side of an equation to be non-negative since on the left we have an arithmetic (that is, non-negative) value of a square root:
`4-x >=0` or `x <=4`

At this point we can see that these two intervals, `x>=4` AND `x<=4` have only one point in common, `x=4`. So, we can just check if `x=4` is a solution. It is, both sides of this equations equal to zero if `x=4`.
This is a legitimate way to conclude this description in this particular case.

However, we might not notice that `x=4` is the only point in common of two conditions for `x`. In this case we should proceed straight to the solution using regular algebraic transformations.

Square both sides of the equation:
`2x-8 = (4-x)^2`

Then
`2x-8 = 16-8x+x^2`
`x^2-10x+24 = 0`
`x_1=6`, `x_2=4`

To no surprise, we have received a solution `x=4` mentioned already above. It satisfies both conditions, `x>=4` and `x<=4`.

The second solution, `x=6`, does not satisfy one of the conditions we started our process with (`x<=4`) and must be discarded.

CHECK of the found solution has already been performed above, no need to repeat it here, but, in general, must always be performed.