How do you solve #4-x=sqrt(x-4)#?
The easy way
So the only solution is
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).
#x^2-9x+20 = 0#
This factors as:
#(x-4)(x-5) = 0#
Then check the solutions:
Note that this happens because numbers have two square roots. When you square both sides of an equation you throw away some information.