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 $16$

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 $t = 16.$
