solving square root and other radical equations

Question

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

^= raised to the power of

Answer 1

Note that we have 5 - x within the brackets.
Since it is to the power of 1/2, this is a square root, which means 5 - x must be more or equal to zero.

Thus, we have 5 - x >= 0, which we can rearrange into x <= 5

(5 - x)^(1/2) = x + 1
5 - x = (x + 1)^2
5 - x = x^2 + 2x + 1
x^2 + 2x + 1 + x - 5 = 0
x^2 + 3x - 4 = 0
(x + (3/2))^2 - 9/4 - 4 = 0
(x + (3/2))^2 - 25/4 = 0
(x + (3/2))^2 = 25/4
x + (3/2) = sqrt(25/4) or = -sqrt(25/4)
x + 3/2 = 5/2 or = -5/2
x = 5/2 - 3/2 or x = -5/2 - 3/2
x = 2/2 or x = -8/2
x = 1 or x = -4

Both answers are less than 5, so they are both acceptable.
HOWEVER, x = 1 assume you are taking a positive square root, whereas x = -4 assume that you are taking a negative square root.

Answer 2

Use a calculator to find the cube root to the nearest hundredth. See Objective 2.

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