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

1 Answer
Jun 24, 2016

Answer:

#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.