<s>[INST] Solve equations $\sqrt{t +9} - \sqrt{t} = 1$ Solve equation:  $\sqrt{t +9} - \sqrt{t} = 1$
I moved - √t to the left side of the equation $\sqrt{t +9} = 1 -\sqrt{t}$
I squared both sides $(\sqrt{t+9})^2  = (1)^2 (\sqrt{t})^2$
Then I got $t + 9 = 1+ t$
Can't figure it out after that point.
The answer is $16$
 [/INST] From $\sqrt{t +9} - \sqrt{t} = 1$ you don't get $\sqrt{t +9} = 1 -\sqrt{t}$ but $$\sqrt{t +9} = 1 +\sqrt{t}.$$
You can solve as follows, using the algebraic identity $(a+b)^2=a^2+2ab+b^2$: 
$$
\begin{eqnarray*}
\sqrt{t+9}-\sqrt{t} &=&1\Leftrightarrow \sqrt{t+9}=1+\sqrt{t}\tag{1} \\
\text{Square both sides of $(1)$} &\Rightarrow &\left( \sqrt{t+9}\right) ^{2}=\left(
1+\sqrt{t}\right) ^{2} \\
\text{Compute and simplify} &\Leftrightarrow &t+9=1+2\sqrt{t}+t\Leftrightarrow 9=1+2\sqrt{t} \\
\text{Simplify} &\Leftrightarrow &9-1=2\sqrt{t}\Leftrightarrow 8=2\sqrt{t} \\
\text{Simplify} &\Leftrightarrow &\frac{8}{2}=\sqrt{t}\Leftrightarrow 4=
\sqrt{t}\tag{2} \\
\text{Square both sides of $(2)$} &\Rightarrow &4^{2}=\left( \sqrt{t}\right) ^{2} \\
&\Leftrightarrow &16=t.\tag{3}
\end{eqnarray*}
$$
Final comment. When we square both sides of an equation we get a new equation which has the same solutions of the original equation, but can have additional solutions. However in this case we got only the solution $t=16$, which is a solution of $(1)$ too.  
ADDED. In your recent question solve the equation $\sqrt{3x-2}+2-x=0$, we get two solutions after squaring 
$$
\begin{equation*}
\sqrt{3x-2}+2-x=0\Rightarrow 3x-2=x^{2}-4x+4\Leftrightarrow x\in \{1,6\}
\end{equation*}
$$
but only $x=6$ is a solution of $\sqrt{3x-2}+2-x=0$, as explained in
this comment by Glen O. 
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