√(1-x) - 1 = x
√(1-x) = x + 1
1 - x = (x + 1)^2
1 - x = x^2 + 2x + 1
0 = x^2 + 3x
x(x + 3) = 0
x = 0, x = -3
2.√(4x^2 + 5x - 2) = 0
4x^2 + 5x - 2 = 0
Using the quadratic formula:
x = [-5 ± √(5^2 - 4 4 -2)] / 8
x = [-5 ± √(25 + 32)] / 8
x = [-5 ± √57] / 8
x = (-5 + √57) / 8 or x = (-5 - √57) / 8
So the solutions are x = (-5 + √57) / 8 and x = (-5 - √57) / 8.
√(1-x) - 1 = x
√(1-x) = x + 1
1 - x = (x + 1)^2
1 - x = x^2 + 2x + 1
0 = x^2 + 3x
x(x + 3) = 0
x = 0, x = -3
2.
√(4x^2 + 5x - 2) = 0
4x^2 + 5x - 2 = 0
Using the quadratic formula:
x = [-5 ± √(5^2 - 4 4 -2)] / 8
x = [-5 ± √(25 + 32)] / 8
x = [-5 ± √57] / 8
x = (-5 + √57) / 8 or x = (-5 - √57) / 8
So the solutions are x = (-5 + √57) / 8 and x = (-5 - √57) / 8.