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
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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
