x - 2√x - 80 = 0

 

There are two ways to solve this:

Method 1:  By u-substitution:

let u = √x
then u² = x

Substitute u² for x and u for √x

u² - 2u - 80 = 0

That factors as:

(u - 10)(u + 8) = 0

u = 10, u = -8

Substitute √x for u

   √x = 10           √x = -8
(√x)² = 10²       (√x)² = (-8)²
    x = 100           x = 64

Check for extraneous solution(s):

     x - 2√x - 80 = 0
 100 - 2√100 - 80 = 0
 100 - 2(10) - 80 = 0
    100 - 20 - 80 = 0
                0 = 0

That's true, so x=10 is a solution.

     x - 2√x - 80 = 0
   64 - 2√64 - 80 = 0
   64 - 2(8) - 80 = 0
     64 - 16 - 80 = 0
              -32 = 0

That's false, so the only solution is x=10



------------------------------
Method 2:  By isolating the radical term and squaring
both sides:

    x - 2√x - 80 = 0
          x - 80 = 2√x
       (x - 80)² = 4(√x)²
x² - 160x + 6400 = 4x 
x² - 164x + 6400 = 0
   (x-100)(x-64) = 0

x = 100, x = 64

Check for extraneous solution(s):

     x - 2√x - 80 = 0
 100 - 2√100 - 80 = 0
 100 - 2(10) - 80 = 0
    100 - 20 - 80 = 0
                0 = 0

That's true, so x=10 is a solution.

     x - 2√x - 80 = 0
   64 - 2√64 - 80 = 0
   64 - 2(8) - 80 = 0
     64 - 16 - 80 = 0
              -32 = 0

That's false, so the only solution is x=10

Edwin