√(x^2-3x+2)+1=√(x-1)+√(x−2)

To solve this equation, we need to isolate the square roots on one side of the equation and then square both sides to eliminate the square roots.

Given equation: √(x^2-3x+2) + 1 = √(x-1) + √(x-2)

Subtract 1 from both sides:
√(x^2-3x+2) = √(x-1) + √(x-2) - 1

Square both sides:
(x^2-3x+2) = (x-1) + 2√(x-1)(x-2) + (x-2) - 2√(x-1)(x-2) + 1

Simplify:
x^2 - 3x + 2 = x - 1 + x - 2 + 1
x^2 - 3x + 2 = 2x - 2

Rearrange the terms:
x^2 - 3x + 2 = 2x - 2
x^2 - 5x + 4 = 0

Now, we have a quadratic equation. We can factor it or use the quadratic formula to find the solutions for x.

Factoring:
(x-4)(x-1) = 0

Setting each factor to zero:
x-4 = 0 or x-1 = 0
x = 4 or x = 1

Therefore, the solutions to the equation are x = 4 and x = 1.