I ran across this equation…
$\sqrt {2x+6}+4=x+3$
Without thinking, I solved for x in the following way:
$\sqrt {2x+6}+4=x+3$
Subtract 4 from both sides.
$\sqrt {2x+6}=x-1$
Square each side.
$2x+6=(x-1)^2$
"FOIL" and rearrange.
$ x^2-4x-5=0 $
Factor.
$(x+1)(x-5)=0$
If you plot the functions you can very clearly see there is only one solution, x=5, which of course stems from the fact that the radical is does not have the plus/minus sign so only half of the "parabola" is considered. But my problem with this is that it should be possible to show algebraically that there is only one solution (other than plugging the numbers in and checking). My thinking is that there is some logical mistake in the way I solved for x, but I can't find it. Moreover, there are of course equations that do intersect twice ($\sqrt {4x+6}+1=x+3$ at $x=\sqrt 2 , -\sqrt 2$ for example), and the exact same steps work in that case.
Any thoughts as to what the differences are between the two cases would be appreciated.
Best Answer
I’m not sure if this is what you mean by see it algebraically, but $\sqrt{2x+6}=x-1\geq 0$ implies $x\geq 1$. This would automatically eliminate $x=-1$.