Answer:
LETTER D (none)
Step-by-step explanation:
√2-x = x-2
(√2-x)² = (x-)²
2-x = x²-4x+4
0 = x²-4x+x+4-2
0 = x²-3x-2
0 = (x-2)(x-1)
0 = x-2 and 0 = x-1
2 = x and 1 = x
Therefore , the roots are 1 and 2
To check the extraneous root , we will substitute the value of x which are 1 and 2
For x = 1
√2-x = x-2
√2-1 = 1-2
√1 = -1
±1 = -1 , it is true
For x = 2
√2-x = x-2
√2-2 = 2-2
√0 = 0
0 = 0 , it is true
Now , we conclude that there are no extraneous root for the solutions
