Question 18028: The worksheet states:
"Solving Equations Containing Radicals. Solve each equation. Be sure to check for extraneous solutions."
Answer by josmiceli(19441) (Show Source):
You can put this solution on YOUR website!
Let w = sqrt( x+10 )
Let z = sqrt( x-6 )
Then w+z = 8
multiply both sides by ( w-z )
( w+z )( w-z ) = 8 ( w-z )
multiplying left side:
w^2 - z^2 = 8( w-z )
w^2 = x+10
z^2 = x-6
substituting:
x+10 - (x-6) = 8( w-z )
16 = 8( w-z )
dividing both sides by 8:
w-z = 2
recall, w+z = 8
w-z = 2
w+z = 8
adding:
2w = 10
w = 5
z = 3
substituting:
sqrt(x-6) = 3
square both sides:
x-6 = 9
x = 15
also:
sqrt(x+10) = 5
square both sides:
x+10 = 25
x = 15
This confirms the result.
x = 15 plugged into the original equation gives the solution:
Note the the positive square roots must be taken.
