I'm trying to find the intersection point of y=x-2 and y=sqrt(x). I tried letting both equations equal each other but kept coming up with the points (4,1) and (1,4) however these aren't right as the answer should be (4,2) but I am unable to get this answer. Help please?

so x-2 = √x

square both sides ...
x^2 - 4x + 4 = x
x^2 - 5x + 4 = 0
(x-1)(x-4) = 0
x = 1 or x = 4

since we squared, both answers have to be verified in the original equation
if x = 1
LS = 1-2 = -1
RS = √1 = 1 ≠ LS

if x=4
LS = 4-2 = 2
RS = √4 = 2

so x = 4 is the only solution

if x = 4,
y = 4-2 = 2

They intersect at (4,2)

look at how Wolfram shows it
http://www.wolframalpha.com/input/?i=x-2+%3D+√x

To find the intersection point of two equations, you need to set them equal to each other and solve for the variables. In this case, we have the equations y = x - 2 and y = sqrt(x).

Let's proceed step by step to find the intersection point:

1. Set the two equations equal to each other:
x - 2 = sqrt(x)

2. Square both sides of the equation to eliminate the square root:
(x - 2)^2 = (sqrt(x))^2
x^2 - 4x + 4 = x

3. Simplify the equation:
x^2 - 4x + 4 - x = 0
x^2 - 5x + 4 = 0

4. Factor the quadratic equation:
(x - 1)(x - 4) = 0

5. Set each factor equal to zero and solve for x:
x - 1 = 0 --> x = 1
x - 4 = 0 --> x = 4

Now, we have two possible values for x: x = 1 and x = 4.

6. Substitute each value of x back into one of the original equations to find the corresponding y-coordinate:

For x = 1:
y = sqrt(1) = 1

For x = 4:
y = sqrt(4) = 2

Therefore, the intersection points are (1, 1) and (4, 2). The point (4, 2) is indeed a valid intersection point for the given equations.

It seems there might have been some confusion with the expected answer. But according to the calculations, both (4, 1) and (1, 4) are not accurate intersection points for the given equations. The correct intersection point is indeed (4, 2).