$\ \sqrt{x+39}-\sqrt{x+7}=4 $ So I tried to solve this problem for x
$\ \sqrt{x+39}-\sqrt{x+7}=4 $
I multiplied both sides ($\ \sqrt{m}\cdot\sqrt{n}=\sqrt{mn} $)
$\ (\sqrt{x+39}-\sqrt{x+7})^2=16 $
$\ (x+39)-2(x^2+46x+273)-(x+7) $
$\ 0x+32+(-2x^2-92x-546) $
$\ -2x^2-92x-514 $
divide the 2 out
$\ x^2+46x+257=-8 $
$\ x^2+46x+265=0 $
Use the quadratic formula (or scientific calculator) and the answers are -6.752 and -39.248. I know the answer is exactly -3. What went wrong?
 A: One can square twice. I prefer to invert and obtain
$$\frac{1}{\sqrt{x+39}-\sqrt{x+7}}=\frac{1}{4},$$
and then by rationalizing the denominator get
$$\frac{\sqrt{x+39}+\sqrt{x+7}}{32}=\frac{1}{4},$$
or equivalently 
$$\sqrt{x+39}+\sqrt{x+7}=8.$$
Now "add" to the original equation, divide by $2$. We get $\sqrt{x+39}=6$, and now it's over. 
Remark: In the original post, the first step used was to square. This to some degree complicates things, the middle term should have been $-2\sqrt{(x+39)(x+7)}$. Rearrangement and squaring now get us to where we want.  
A: $$
\sqrt{x + 39} - \sqrt{x + 7} = 4 \quad \Rightarrow \quad (\sqrt{x + 39})^2 = (\sqrt{x + 7} + 4)^2 \quad \Rightarrow 
$$
$$
x + 39 = x + 7 + 8\sqrt{x + 7} + 16 \quad \Rightarrow \quad 8\sqrt{x + 7} = 16 \quad \Rightarrow \quad x = -3
$$
A: You made two mistakes when squaring. The correct equation is
$$(x + 39) - 2\sqrt{(x+39)(x+7)} + (x + 7) = 16$$
This can be rearranged as
$$2x + 30 = 2\sqrt{(x+39)(x+7)}$$
Dividing by 2:
$$x + 15 = \sqrt{(x+39)(x+7)}$$
Square again
$$x^2 + 30x + 225  = (x+39)(x + 7)$$
$$= x^2 + 46x + 273$$
So your equation is the same as
$$30x + 225 = 46x + 273$$
So $16x = -48$ and therefore $x = -3$.
When squaring equations always plug in your answer to make sure it's the actual solution. And in this case it is.
