First, we can square both sides of the equation to eliminate the fractional exponent while keeping the equation balanced:
((5 - x)^(1/2))^2 = (x + 1)^2
((5 - x)^1 = x^2 + 2x + 1
5 - x = x^2 + 2x + 1
Next, we can convert the equation into standard form:
5 - x - color(red)(5) + color(blue)(x) = x^2 + 2x + 1 - color(red)(5) + color(blue)(x)
5 - color(red)(5) - x + color(blue)(x) = x^2 + 2x + color(blue)(x) + 1 - color(red)(5)
0 - 0 = x^2 + 3x - 4
0 = x^2 + 3x - 4
x^2 + 3x - 4 = 0
Now, we can use the quadratic equation to solve this problem:
The quadratic formula states:
For color(red)(a)x^2 + color(blue)(b)x + color(green)(c) = 0, the values of x which are the solutions to the equation are given by:
x = (-color(blue)(b) +- sqrt(color(blue)(b)^2 - (4color(red)(a)color(green)(c))))/(2 * color(red)(a))
Substituting:
color(red)(1) for color(red)(a)
color(blue)(3) for color(blue)(b)
color(green)(-4) for color(green)(c) gives:
x = (-color(blue)(3) +- sqrt(color(blue)(3)^2 - (4 * color(red)(1) * color(green)(-4))))/(2 * color(red)(1))
x = (-color(blue)(3) +- sqrt(9 - (-16)))/2
x = (-color(blue)(3) - sqrt(9 - (-16)))/2 and x = (-color(blue)(3) + sqrt(9 - (-16)))/2
x = (-color(blue)(3) - sqrt(9 + 16))/2 and x = (-color(blue)(3) + sqrt(9 + 16))/2
x = (-color(blue)(3) - sqrt(25))/2 and x = (-color(blue)(3) + sqrt(25))/2
x = (-color(blue)(3) - 5)/2 and x = (-color(blue)(3) + 5)/2
x = -8/2 and x = 2/2
x = -4 and x = 1
Substituting both solutions into the original equation gives:
(5 - (-4))^(1/2) = -4 + 1 and (5 - 1)^(1/2) = 1 + 1
(5 + 4)^(1/2) = -3 and (4)^(1/2) = 2
(9)^(1/2) = -3 and 2 = 2
3 != -3 and 2 = 2
Therefore the solution x = -4 is extraneous.
