Isolate the radical by adding 
to both sides:
Now square both sides:
On the left side, to square a square root takes
away the radical and the exponent. We write the
right side as 
Use FOIL on the right side
See if you can go from there to:
Then we factor the left side:
Use the zero factor principle,
and set each factor = 0:
gives 
gives 
But we must check both solutions in the
original equations, because in radical
equations, there may be extraneous (bogus)
solutions.
Checking in the original equation:
That is false, so is not a solution
to the original equation.
Checking in the original equation:
That is true, so is the only solution
to the original equation.
---------------------
If you look at the graph of the left side of
that is, plot 
you get this curve:
and if you plot the right side of
that is, plot 
on the same graph you see that the only
solution is the x-coordinate of the point
where they intersect.
They intersect only at (2,3). The x-coordinate
of that point is 2, so is the only
solution to the system. But we wonder why we
got the extraneous solution.
We now notice that the curve of the left
side is really part of a slanted parabola,
and if we were to draw in the rest of the
parabola, that is, the part in light blue
below:
we can see that the other part of that parabola
does indeed intersect the graph of the right side,
(the green line) at the point (6,3). But we
discard because the curve of the left side
of the original equation does not include the light
blue part of the parabola.
Edwin
