When solving the given equation √2x + 1 = √√√x+1 - 2, if we first square both sides, we get:
2x + 1 = (√√√x+1 - 2)²
Simplifying the right-hand side using the formula for the square of a binomial, we get:
2x + 1 = (√√√x+1)² - 4(√√√x+1) + 4
2x + 1 = (√√√x+1)² - 4(√√√x+1) + 5
Substituting u = √√√x+1, we can rewrite this equation as:
2u⁴ - 4u² + 1 = 0
This is a quartic equation, which can be difficult to solve algebraically. However, we can notice that u = 1 is a solution of this equation, since:
2(1)⁴ - 4(1)² + 1 = 0
Therefore, u = √√√x+1 = 1, which implies that √√x+1 = 1, and hence that √x+1 = 1² = 1. Squaring both sides, we get:
x + 1 = 1
Subtracting 1 from both sides, we get:
x = 0
However, we need to check whether this solution is valid, since we squared both sides of the equation at the beginning. When we square both sides of an equation, we can introduce extraneous solutions that are not valid solutions of the original equation. To check whether x = 0 is an extraneous solution, we need to substitute it back into the original equation and see if it satisfies the equation.
Substituting x = 0 into the original equation, we get:
√2(0) + 1 = √√√(0)+1 - 2
1 = √√1 - 2
1 = √-1
The square root of a negative number is not a real number, so x = 0 is not a valid solution of the original equation. Therefore, the extraneous solution that arises when squaring both sides of the equation is x = 0.
Therefore, the correct answer is x = 0.
