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

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sciencesolve | Teacher | (Level 3) Educator Emeritus

Posted on

You need to convert the rational power `1/2` into a radical, such that:

`(x + 1)^(1/2) = sqrt(x + 1)`

Substituting `sqrt(x + 1)` for `(x + 1)^(1/2)` in equation, yields:

`sqrt(x + 1) = 5 - x`

You need to remove the square root, hence, you need to raise to square to the left and right sides, such that:

`x + 1 = (5 - x)^2 => x + 1 = 5^2 - 2*5*x + (-x)^2`

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

You need to move all terms to one side, such that:

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

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

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

You need to group the terms such that:

`(x^2 - 3x) - (8x - 24) = 0`

You need to factor out x in the first group such that:

`x(x - 3) - (8x - 24) = 0`

You need to factor out 8 in the second group, such that:

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

Factoring out `(x - 3)` yields:

`(x - 3)(x - 8) = 0 => {(x - 3 = 0),(x - 8 = 0):}`

`{(x = 3),(x = 8):}`

You need to test the values x = 3 and x = 8 in equation, such that:

`sqrt(3 + 1) = 5 - 3 => sqrt 4 = 2 => +-2 = +2 => 2 = 2`

`sqrt(8 + 1) = 5 - 8 => sqrt 9 = -3 => +-3=-3 => -3 = 3`

Hence, evaluating the solutions to the given equation yields `x = 3` and `x = 8` .

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giorgiana1976 | College Teacher | (Level 3) Valedictorian

Posted on

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.

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