# What is x if (x+1)^1/2=5-x ?

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We have to find x using the equation (x+1)^1/2 = 5 - x

(x+1)^1/2 = 5 - x

square both the sides

x+1 = 5^2 + x^2 - 10x

=> x + 1 = 25 + x^2 - 10x

=> x^2 - 11x + 24 = 0

=> x^2 - 8x - 3x + 24 = 0

=> (x - 8)(x -3 ) =0

From x - 8 = 0, we get x = 8

And x - 3 = 0 gives x = 3

**Therefore x can take the values 3 and 8.**

We'll impose the constraint of existence of the square root:

x+1 >= 0

x >= -1

Now, we'll solve the equation by raising the square both sides:

[sqrt(x+1)]^2 = (5-x)^2

We'll expand the square form the right side:

x + 1 = 25 - 10x + x^2

We'll subtract x+1 both sides:

x^2 - 10x + 25 - x - 1 =0

We'll combine like terms:

x^2 - 11x + 24 = 0

We'll apply the quadratic formula:

x1 = [11+sqrt(121 - 96)]/2

x1 = (11+5)/2

**x1 = 8**

**x2 = 3**

**Since both solutions are in the interval of admissible values, they are accepted.**