x = 4 is an extraneous solution.
√(5-x) -√x=1
√(5-x) = 1 + √x
Squaring both sides of the equations, we get
(√(5-x))² = (1 + √x)²
5 - x = 1 + x + 2√x
4 - 2x = 2√x
Squaring again both sides of the equations, we get
(4 - 2x)² = (2√x)²
16 + 4x² - 16x = 4x
4x² - 20x + 16 = 0 ...(1)
By middle term splitting equation (1) can be written as
4x² - 16x - 4x + 16 = 0
4x(x - 4) -4(x - 4) = 0
(x - 4) (4x - 4) = 0
4x - 4 = 0 and x - 4 = 0
x = 1 and 4
Check for Extraneous solution by putting value of x in the original equation -
Let x = 1
√(5-1) -√1=1
√4 - 1 = 1
2 -1 = 1
1 = 1
Since, LHS = RHS, so, x = 1 is not an extraneous solution
Let x = 4
√(5-4) -√4=1
√1 - 2 = 1
1 - 2 = 0
-1 = 0
which is a contradiction, so, x = 4 is an extraneous solution.
Know more about extraneous solution:
brainly.com/question/1511979
#SPJ4
