1. ## Solve Equation

Solve:

$x-{\sqrt{x+4}}=2$

2. This equation is somewhat ambiguous.

$x-\sqrt{x+4}=2$ becomes $x^2-x+4=4$ which becomes $x^2=x$ 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?

3. Originally Posted by acevipa
Solve:

$x-{\sqrt{x+4}}=2$
Isolate the square root:
$-\sqrt{x + 4} = -x + 2$

Now square both sides:
$x + 4 = (-x + 2)^2 = x^2 - 4x + 4$

$0 = x^2 - 5x$

$0 = x(x - 5)$

So x = 0 or x = 5. Always check these in your original equation. I get that only the x = 5 is a solution.

-Dan

4. Originally Posted by mrbuttersworth
This equation is somewhat ambiguous.

$x-\sqrt{x+4}=2$ becomes $x^2-x+4=4$ which becomes $x^2=x$ 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?
$(x - \sqrt{x + 4})^2 \neq x^2 - (\sqrt{x + 4})^2$

Make sure you FOIL these out so that you can see how they square. The general rule is
$(a \pm b)^2 = a^2 \pm 2ab + b^2$

-Dan

5. Originally Posted by acevipa
Solve:

$x-{\sqrt{x+4}}=2$
Before solving equations like this one, we require certain conditions:
• $x\ge-4.$
• $x\ge2.$

Hence, we require that $x\ge2.$ Now, by makin' the straightforward calculations we get $x^2-5x=0,$ which is Dan's result. From here $x=0$ or $x=5,$ but the answer must lie on $[2,\infty)$ (our requirement), so $x=5.$