Find Intersection of -sqrt(x) & x-6 Functions

[zeion]

Homework Statement

I need to find the point at which these functions intersect:
y = -sqrt(x)
y = x - 6

Homework Equations

The Attempt at a Solution

I set them to equate:
-sqrt(x) = x - 6
-sqrt(x) - x + 6 = 0

Now how do I find the root?

 

[icystrike]

equate them tgt , $$\sqrt{x}=6-x$$ .
Now square both sides , x=36+x²-12x ... continue from here.

 

[zeion]

Do I move them all to one side and get x^2 -13x + 36 = 0?

 

[Mark44]

Yes. Now factor.

Be sure to check each solution in the original equation, though. When you square both sides of an equation, extraneous solutions are sometimes introduced, values that are not solutions of your original equation.

 

[ideasrule]

Alternatively, you can say u=sqrt(x) and rewrite the equation as -u-u^2+6=0

 

[zeion]

Okay nice.
So I can't use x = 9 because -sqrt(9) not= 9-6, correct?

 

[zeion]

Ah okay, that u substitution seems easier:
Let u = sqrt(x), then
u^2 + u - 6 = 0
(u+3)(u-2) = 0
u = -3, u = 2
sqrt(x) = -3, sqrt(x) = 2
x = 9, x = 4

 

[zeion]

Thanks guys!

 

[Mark44]

Ah okay, that u substitution seems easier:
Let u = sqrt(x), then
u^2 + u - 6 = 0
(u+3)(u-2) = 0
u = -3, u = 2
sqrt(x) = -3, sqrt(x) = 2
x = 9, x = 4

But x = 9 is not a solution of sqrt(x)= 6 -x, which is what you started with. x = 9 is an extraneous solution that I warned you of.