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 $16$
 A: Hint. Let $x=\sqrt{t}$ and $y=\sqrt{t+9}$. Then 
$$
y-x=1\implies y^2-x^2=y+x\implies 9=y+x
$$
A: 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. 
A: $$\sqrt{t +9} - \sqrt{t} = 1$$
Multiplying by $\sqrt{t +9} + \sqrt{t}$ you get
$$9=\sqrt{t +9} +\sqrt{t} $$
Now adding
$$\sqrt{t +9} + \sqrt{t} =9$$
$$\sqrt{t +9} - \sqrt{t} = 1$$
you get 
$$\sqrt{t+9}=5 \Rightarrow t=25-9 $$
A: after squaring both sides you get:
$$t+9=1+2\sqrt{t}+t\implies4=\sqrt{t} \implies t=16$$
A: The solution of such a problem such a problem lies in finding a number for which the equation holds.  Suppose you didn't know the answer was 16, how might you solve the problem?
Well, you have two square roots in the problem $\sqrt{t +9}$ and $\sqrt t$.  Their squares differ by 9.  Thus, if you can find two square numbers which differ by 9, then the larger square number will equal (t +9) and the smaller square number will equal t.  The set of square numbers, as you might know, consists of the sequence (1, 4, 9, 16, 25, ...).  25 and 16 differ by 9, and 16 is the smaller.  Thus, we have found 16 as the solution to this equation.
A: An often overlooked fact is that
$$ \sqrt{t^2}=\left|t\right|$$
Call me paranoid but here's how I would solve this
$$\sqrt{t +9} - \sqrt{t} = 1$$
$$\sqrt{t +9}  = \sqrt{t}+1$$
$$\left|t +9\right|  = \left(\sqrt{t}+1\right)\left(\sqrt{t}+1\right)$$
$$\left|t +9\right|  = \left|t\right|+2\sqrt{t}+1$$
Since the original equation has $\sqrt{t}$, then we know that $t\geq 0$ and we can safely remove the absolute value bars. So now we have
$$t +9= t+2\sqrt{t}+1$$
$$9= 2\sqrt{t}+1$$
$$8= 2\sqrt{t}$$
$$4= \sqrt{t}$$
$$\left|t\right|=4^2$$
$$t=16$$
A: Consinder the function
$f(t) = \sqrt{t+9}-\sqrt{t}-1$, $t\geqslant 0$.
We have $f'(t) = \dfrac{1}{2\sqrt{t+9}} - \dfrac{1}{2\sqrt{t}} < 0$.
Therefore $f$ is a dececreasing function on the  $(0, +\infty)$.
Another way, $t = 16$ is a solution of the given equation.
Therefore the given equation has only solution $t = 16.$
