Simple, find-intersection task by calculation. How do I know only one is valid? 
Possible Duplicate:
Is there a name for this strange solution to a quadratic equation involving a square root? 


As you can see, they only cross once. By setting
$3-x=\sqrt{15-x}$
I manage somehow to find two possible values for x.
$3-x=(15-x)^2$
$(3-x)=(15-x)^{\frac{1}{2}}$
$(3-x)^2=((15-x)^{\frac{1}{2}})^2$
$9-6x+x^2=15-x$
$x^2-5x-6=0$
$x_1=6$ and $x_2=-1$
$y_1=3-6=-3$ and $y_2=3-(-1)=4.$
$(6,-3)$ and $(-1,4)$
Only $(-1,4)$ is a real intersection.
How is that this method of calculating the intersection allows me to find this false positive if you will? 
 A: You started with the equation 
$$3-x=\sqrt{15-x},$$
and then squared both sides, obtaining the equation
$$(3-x)^2=15-x.$$
Now suppose that you had started from the equation
$$3-x=-\sqrt{15-x},$$
and squared both sides. Again you would get
$$(3-x)^2=15-x.$$
How is the poor equation $(3-x)^2=15-x$ to know whether it "comes from" $3-x=\sqrt{15-x}$ or from $3-x=-\sqrt{15-x}\;$?  It can't.  So it has to play it safe and give you all the solutions of $3-x=\sqrt{15-x}$, and also all the solutions of $3-x=-\sqrt{15-x}$.
And, as you can readily check, the extraneous solution that you got is indeed the solution of the equation $3-x=-\sqrt{15-x}$.
Geometric interpretation: The curve $y=\sqrt{15-x}$ is half of a parabola.  When you square both sides of $y=\sqrt{x-5}$, you get $y^2=x-5$, which is the full parabola.  So when you squared both sides, the algebra decided that you wanted the intersection points of the line with the full parabola.  And that's exactly what the algebra produced. By looking at the original picture, and the two candidates for solution, you can tell instantly which one is not valid.
A: When you square both sides, you often introduce extraneous solutions.  Squaring is a non-reversible operation, which is why you create them.  In your case, plugging $6$ in to the left gives $3-6=-3$, while plugging it in to the right gives $\sqrt{15-6}=3$.  As $(-3)^2=3^2$, this "solution" appears.
