Solve:
$\displaystyle x-{\sqrt{x+4}}=2$
This equation is somewhat ambiguous.
$\displaystyle x-\sqrt{x+4}=2$ becomes $\displaystyle x^2-x+4=4$ which becomes $\displaystyle x^2=x$ so therefore, x must be equal to 1 or 0
that's the way I would have done it. This isn't certain though. Can we get clarification?
Isolate the square root:
$\displaystyle -\sqrt{x + 4} = -x + 2$
Now square both sides:
$\displaystyle x + 4 = (-x + 2)^2 = x^2 - 4x + 4$
$\displaystyle 0 = x^2 - 5x$
$\displaystyle 0 = x(x - 5)$
So x = 0 or x = 5. Always check these in your original equation. I get that only the x = 5 is a solution.
-Dan
Before solving equations like this one, we require certain conditions:
- $\displaystyle x\ge-4.$
- $\displaystyle x\ge2.$
Hence, we require that $\displaystyle x\ge2.$ Now, by makin' the straightforward calculations we get $\displaystyle x^2-5x=0,$ which is Dan's result. From here $\displaystyle x=0$ or $\displaystyle x=5,$ but the answer must lie on $\displaystyle [2,\infty)$ (our requirement), so $\displaystyle x=5.$