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