How do you solve #sqrt(2x+20)+2=x#?

1 Answer
Aug 8, 2015

Answer:

# x = 8 #

Explanation:

The trick to solve these kind of problems is to take all the non-radical expressions on one side, and then square both sides to remove the radical. Let us try out our strategy for this particular problem.

# sqrt( 2x + 20 ) + 2 = x #
# => sqrt( 2x + 20 ) = x - 2 #
# => 2x + 20 = ( x - 2 )^2 # (Squaring both sides)
# => 2x + 20 = x^2 - 4x + 4 #
# => x^2 - 6x - 16 = 0 #
# => x^2 - 8x + 2x - 16 = 0 #
# => x ( x - 8 ) +2 ( x - 8 ) = 0 #
# => ( x + 2 ) ( x - 8 ) = 0 #
# => x = -2 # or # x = 8 #

Now, we need to check the validity of our solution.
First, we check # x = 8 #. Putting the value in the LHS, we get #8#, so it is a solution.
Next we check # x = -2 #. Putting the value in the LHS, we get #6#. But the RHS# = x = -2 #. So # -2 # is not a valid solution of the equation. These kind of roots are called extraneous roots in literature.