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 √2x+20+2=x
=> sqrt( 2x + 20 ) = x - 2 ⇒√2x+20=x−2
=> 2x + 20 = ( x - 2 )^2 ⇒2x+20=(x−2)2 (Squaring both sides)
=> 2x + 20 = x^2 - 4x + 4 ⇒2x+20=x2−4x+4
=> x^2 - 6x - 16 = 0 ⇒x2−6x−16=0
=> x^2 - 8x + 2x - 16 = 0 ⇒x2−8x+2x−16=0
=> x ( x - 8 ) +2 ( x - 8 ) = 0 ⇒x(x−8)+2(x−8)=0
=> ( x + 2 ) ( x - 8 ) = 0 ⇒(x+2)(x−8)=0
=> x = -2 ⇒x=−2 or x = 8 x=8
Now, we need to check the validity of our solution.
First, we check x = 8 x=8. Putting the value in the LHS, we get 88, so it is a solution.
Next we check x = -2 x=−2. Putting the value in the LHS, we get 66. But the RHS = x = -2 =x=−2. So -2 −2 is not a valid solution of the equation. These kind of roots are called extraneous roots in literature.
