Radical equation solve $\sqrt{3x+7}-\sqrt{x+2}=1$. Cannot arrive at solution $x=-2$ I am to solve $\sqrt{3x+7}-\sqrt{x+2}=1$ and the solution is provided as -2.
Since this is a radical equation with 2 radicals, I followed suggested textbook steps of isolating each radical and squaring:
$\sqrt{3x+7}-\sqrt{x+2}=1$ 
$(3x+7=(1-\sqrt{x+2})^2$ # square both sides
(Use perfect square formula on right hand side $a^2-2ab+b^2$)
$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$ # lhs radical is removed, rhs use perfect square formula
$3x+7=1+2(\sqrt{x+2})+x+2$ # simplify
$3x+7=x+3+2\sqrt{x+2}$ # keep simplifying
$2x+4=2\sqrt{x+2}$ # simplify across both sides
$(2x+4)^2=(2\sqrt{x+2})^2$
$4x^2+16x+16=4(x+2)$ # now that radical on rhs is isolated, square both sides again
$4x^2+12x+14=0$ # a quadratic formula I can use to solve for x
For use int he quadratic function, my parameters are: a=4, b=12 and c=14:
$x=\frac{-12\pm\sqrt{12^2-(4)(4)(14)}}{2(4)}$
$x=\frac{-12\pm{\sqrt{(144-224)}}}{8}$
$x=\frac{-12\pm{\sqrt{-80}}}{8}$
$x=\frac{-12\pm{i\sqrt{16}*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{4i*i\sqrt{5}}}{8}$
$x=\frac{-12\pm{-4\sqrt{5}}}{8}$ #since $4i*i\sqrt{5}$ and i^2 is -1
This is as far as I get:
$\frac{-12}{8}\pm\frac{4\sqrt{5}}{8}$
I must have gone of course somewhere further up since the solution is provided as x=-2. 
How can I arrive at -2?
 A: Hint: Writing your equation in the form
$$\sqrt{3x+7}=1+\sqrt{x+2}$$
squaring gives
$$3x+7=1+x+2+2\sqrt{x+2}$$ so
$$x+2=\sqrt{x+2}$$ squaring again:
$$(x+2)^2=x+2$$
Can you finish?
A: Let $\sqrt{3x+7}=a,\sqrt{x+2}=b$
$\implies a,b\ge0$ and $a-b=1$
and $a^2-3b^2=1$
Or $(b+1)^2-3b^2=1$
A: Your first step is wrong. It should be $$\sqrt{3x+7}-\sqrt{x+2}=1\implies\sqrt{3x+7}=1+\sqrt{x+2}$$ so we have $$3x+7=(1+\sqrt{x+2})^2$$ from which I think you can continue.

Note that as a check to your textbook solution, at $x=-2$, we get $$\sqrt{3(-2)+7}-\sqrt{-2+2}$$ which is indeed equal to $1$.
A: 
$\sqrt{3x+7}-\sqrt{x+2}=1$
$3x+7=(1 \color{red}{\mathbf{ \,-\, }}\sqrt{x+2})^2$ # square both sides

You want: $3x+7=(1 \color{blue}{\mathbf{ \, + \,}}\sqrt{x+2})^2$

Note: not only $x=-2$ solves this equation, also $x=-1$.
A: In addition to the possible typo $\sqrt{3x+7}=1+\sqrt{x+2}$ not $1-\sqrt{x+2}$ in the RHS, you made an arithmetic error later.
From $4x^2+16x+16=4(x+2)$, you should get $4x^2+12x+8=0$, not $4x^2+12x+14=0$.
A: There are actually two solutions: $x = -1; x = -2$ when you continue with the ``Can you finish?'' step of Dr. S.
A: The big error is that $4x^2+16x+16=4(x+2)$ is the same as $4x^2+12x+8=0.$ You somehow got $4x^2+12x+14=0.$ Did you treat $4(x+2)$ as the same as $4x+2?$ The equation $4x^2+12x+8=0$ has $x=-1$ and $x=-2$ as roots.
There's an earlier error where you write: $3x+7=(1-\sqrt{x+2})^2.$ The right side should be $(1+\sqrt{x+2})^2,$ but your later expansion somehow yields the correct value - so two errors led to a correct expression.
It's easier, when you have $2x+4=2\sqrt{x+2},$  if you divide by $2$ before squaring, and get: $x+2=\sqrt{x+2}.$
One quick way to simplify it from the start is to set $y=x+2.$ Then $3y+1=3x+7.$ So you have a slightly simpler equation:
$$\sqrt{3y+1}-\sqrt{y}=1\\
\sqrt{3y+1}=1+\sqrt{y}\\
3y+1 = 1+2\sqrt{y}+y\\
2y=2\sqrt{y}\\
y=\sqrt{y}\\
y^2=y\\
y=0,1$$
You have to go back and check each $y$ in the original equation, then take $x=y-2$ for each solution $y.$
A: Let $x+2=y$, then:
$\sqrt {3y+1}=\sqrt y +1$
squaring both sides we get:
$3y+1=y+1+2\sqrt y$
⇒ $y=\sqrt y $ ⇒ $y^2-y=y(y-1)=0$
that gives:
$y=x+2=0$  ⇒ $x=-2$
$y-1=0$  ⇒ $y=x+2=1$  ⇒ $x=-1$
A: This line
$$3x+7=1^2-2(1)(-\sqrt{x+2})+x+2$$
should be
$$3x+7=1^2-2(1)(\sqrt{x+2})+x+2$$
You put in one too many minuses.

This is my solution.
\begin{align}
   \sqrt{3x+7} - \sqrt{x+2} &= 1 \\
   (\sqrt{3x+7} + \sqrt{x+2})(\sqrt{3x+7} - \sqrt{x+2})
   &= (\sqrt{3x+7} + \sqrt{x+2}) \\
   (3x+7) - (x+2) &= \sqrt{3x+7} + \sqrt{x+2} \\
   \sqrt{3x+7} + \sqrt{x+2} &= 2x+5 \\
\hline
   (\sqrt{3x+7} + \sqrt{x+2}) - (\sqrt{3x+7} - \sqrt{x+2})
   &= (2x+5) - 1 \\
   2\sqrt{x+2} = 2x+4 \\
   \sqrt{x+2} = x+2 \\
   x+2 &= x^2+4x+4 \\
   x^2 + 3x + 2 &= 0 \\
   (x+1)(x+2) &= 0 \\
   x &\in \{-1, -2\}
\end{align}
