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

Question

1644 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-10-17T09:08:11

#x = 4#

The easy way

#sqrt(x-4)# only takes Real values when #x >= 4# and then #sqrt(x-4) >= 0#.

When #x >= 4# then #4 - x <= 0#

So the only solution is #x = 4#

The normal way

Square both sides of the equation to get:

#16-8x+x^2 = (4-x)^2 = (sqrt(x-4))^2 = x - 4#

Note that squaring both sides of an equation may introduce spurious solutions (as it does in this case).

Subtract #x - 4# from both ends to get:

#x^2-9x+20 = 0#

This factors as:

#(x-4)(x-5) = 0#

So #x = 4# or #x = 5#.

Then check the solutions:

If #x = 4# then #4 - x = 0# and #sqrt(x-4) = 0#. So #x=4# is a solution of the original equation.

If #x = 5# then #4 - x = -1# but #sqrt(x - 4) = 1#. So #x=5# is not a solution of the original equation.

Note that this happens because numbers have two square roots. When you square both sides of an equation you throw away some information.