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

Jun 24, 2016

$x = 4$

Explanation:

Start with a condition for existing of a square root on the left side of an equation:
$2 x - 8 \ge 0$ or $x \ge 4$
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:
$4 - x \ge 0$ or $x \le 4$

At this point we can see that these two intervals, $x \ge 4$ AND $x \le 4$ 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:
$2 x - 8 = {\left(4 - x\right)}^{2}$

Then
$2 x - 8 = 16 - 8 x + {x}^{2}$
${x}^{2} - 10 x + 24 = 0$
${x}_{1} = 6$, ${x}_{2} = 4$

To no surprise, we have received a solution $x = 4$ mentioned already above. It satisfies both conditions, $x \ge 4$ and $x \le 4$.

The second solution, $x = 6$, does not satisfy one of the conditions we started our process with ($x \le 4$) 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.