System Confusion?

Question

y = - sqrt(x-6)
y = (x-3)^2 + y^2 = 4

Kinda confused? So its the negative square root of the (x-6), when I square it to solve for the second equation, does the square root also cancel out or just the negative in front?

Answer 1

Actaully both cancel out.

Answer 2

y = - sqrt(x-6)
y = (x-3)^2 + y^2 = 4

y^2 = - sqrt(x-6) * - sqrt(x-6) = + (x-6)
A negative times a negative is a positive. So yes, the negatives cancel out. The sign is positive. So now we multiply the roots to give simply x-6.

y = (x-3)^2 + (- sqrt(x-6))^2 = 4
y = (x-3)^2 + x-6 = 4

I'm not sure where you want to go from here, are you solving for X? Just incase:
(x-3)^2 + x - 6 = 4
x^2 - 6x + 9 + x -6 = 4
x^2 - 5x - 1 = 0
The quadratic formula yields:
x = +(root(29)+5)/2 and -(root(29)+5)/2
x = +5.19 and -5.19

Hope this helps!

Answer 3

y = - sqrt(x-6) ----------------eqn1
(x-3)^2 + y^2 = 4 -----------eqn2

>>>
(x-3)^2 + [- sqrt(x-6) ]^2 = 4
(x^2 -6x+9 ) + (x-6) = 4
x^2 -5x +3 = 4
x^2 -5x -1 = 0
<< no real roots only complex since discriminate <0

when you square a square root, the sign does not make
any difference

Answer 4

If you square -sqrt(x - 6), you'll get x - 6. When you square negative one, you get positive one. A negative value squares to a positive value. Only imaginary values square to negative values.