Can square roots be negative? 1955 AHMSE Problem 20 asks when $\sqrt{25 - t^2} + 5 =0.$
I know square root of real numbers cannot be negative. So t cannot be real.
But I don't know whether imaginary numbers' square root can be negative or not. I think square roots can never be negative. Also, I don't think we can classify imaginary numbers as positive or negative.
Can square roots of imaginary numbers be negative? 
 A: Read carefully the first paragraph of what wikipedia has to say about square roots, noting the difference between the definitions of "square root" and "principal square root":
https://en.wikipedia.org/wiki/Square_root
$\sqrt{x} \ $ doesn't mean "square root of x". $\sqrt{x} \ $ means, by definition, $\mathit{the \ principal \ square \ root \ }$ of x, which means, $\mathit{the \ square \ root \ of \ x \ with \ positive \ real \ part }$ . Note that this definition can be applied to complex numbers also, e.g. for the following A-Level question:

if we did not require the principal square root to have a positive real part, then the answer would be: Solutions are 7-3i and -7+3i, whereas given the definition of principal square root, which is standard convention, we must throw away -7+3i, but keep 7-3i as a solution.
4 and −4 are square roots of 16.
However, $\sqrt{16} = 4$, $\mathit{not} \pm 4$.
For further evidence in favour of my understanding of the conventional meaning of $\sqrt{} \ \ $, In Rudin's PMA, Theorem 1.21 states:
For every real $x > 0$ and every integer $n > 0$ there is one and only one $\mathit{positive}$ real y such that $y^{n} = x$. The number y is written $\sqrt[n]{x} \ $ or $ \ x^{1/n}$.
Funnily enough, as pointed out by LegionMammal978 in the comments, Rudin left out the fact that y is positive in his book.
Going back to the original questions,
"I know square root of real numbers cannot be negative. So t cannot be real."
I agree.
"But I don't know whether imaginary numbers' square root can be negative or not."
All complex numbers have two square roots (though they may be repeats). An imaginary number like $9i$ will be have two complex square roots. This is true for complex numbers also- they too have two complex square roots. 
I actually think what you meant to ask here was, "can we square root a $\mathit{negative}$ number?" In the complex realm, yes, but we have to be careful if we are calculating the principal square root and are using the $\sqrt{} \ \ $ sign. see:
https://en.wikipedia.org/wiki/Imaginary_unit#Proper_use
for why and when we have to be careful.
-4 has two complex square roots, namely $2i$ and $-2i$. Since the real part of both of these square roots is 0, either both or neither are considered principal square roots, depending on if you define the principal square root as having non-negative real part, or strictly positive real part. To be honest I'm not sure which one is "accepted/standard convention", but if I had to guess, I would say the latter option is correct i.e. both are principal square roots of -4. I'm open to confirming whether or not my suspicions of the "accepted/standard convention" are correct or incorrect based on as long as that someone has sources to back up their claim.
If you ask if a number is negative, then in this context you are implicitly implying the number is real and negative, or integer and negative etc, because only real/integer/etc numbers can be negative. Complex numbers with imaginary part not equal to 0 cannot be negative in and of themselves, however, their $\mathit{ \ real \ / \ imaginary \ parts \ }$ can be negative.
"I think square roots can never be negative."
Square roots of real numbers can be negative. -4 is a square root of 16. But remember: $\sqrt{x}$ means the $\mathit{principal}$ square root of x, not just the square root of x. You would be correct to say, "the principal square root of a real number cannot be negative"
"Also, I don't think we can classify imaginary numbers as positive or negative."
Correct. But like I said you $\mathit{can}$ talk about their $\mathit{ \ real \ / \ imaginary \ parts \ }$
A: Did someone read the original question?  Since
$$\sqrt{25-t^2}=-5\Rightarrow 25-t^2=25\iff t=0$$
and zero is no solution to the original equation, option (a) is the right answer.
A: We get from your equation $$\sqrt{25-t^2}=-5$$ since $$\sqrt{25-t^2}\geq 0$$ for $$25-t^2\geq 0$$ so we get no solution.
A: All numbers have two square roots.  By convention the positive square root is assumed to be the square root for positive reals.  However in your example $t=0$ is O.K., since $\sqrt{25}=\pm 5$.
A: You can use the third binomial formula to see that $t=0$ is no solution.
$$\sqrt{25 - t^2} + 5 =0$$
$$\sqrt{25 - t^2} =- 5 $$
$$\sqrt{5 - t}\cdot \sqrt{5+t} =- 5 $$
$$\sqrt{5}\cdot \sqrt{5} =- 5 $$
$$5 =- 5 \qquad \color{red}{\times}$$
A: Let's begin with $\sqrt{25+t^2}+5=0\implies \sqrt{25+t^2}=-5.$ If we square both sides, we get $25+t^2=25\implies t^2=25-25=0\implies t=0.$ Zero is real so $t$ is real.
A: 
Can square roots of imaginary numbers be negative? 

No -- the only thing that can conceivably have a negative number as a square root is the square of that negative number, which is positive real rather than imaginary.
As other answers already point out, the negative number can be considered a square root of its positive square, but is not the principal square root that is usually notated with the $\sqrt{\phantom{z}}$ sign.
The square root of a purely imaginary number is a complex number on one of the lines at 45° angles to the real and imaginary axes. For example,
$$ \sqrt{2i} = 1 + i $$
