A biquadratic equation to have at least two real roots Which of the equations have at least two real roots?
\begin{aligned}
x^4-5x^2-36 & = 0 & (1) \\
x^4-13x^2+36 & = 0 & (2) \\
4x^4-10x^2+25 & = 0 & (3) 
\end{aligned}
I wasn't able to notice something clever, so I solved each of the equations. The first one has $2$ real roots, the second one $4$ real roots and the last one does not have real roots. I am pretty sure that the idea behind the problem wasn't solving each of the equations. What can we note to help us solve it faster?

There is a test that can be made for biquadratics without the need for square-roots that is akin to a test for the existence of real zeroes of a quadratic polynomial (there is an evident connection with the Rule of Signs that Gregory discusses).  There, "completing the square" produces
$$ u^2 \ + \ Bu \ + \ C \ \ = \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ + \ \left(C \ - \ \frac{B^2}{4} \right) \ \ = \ \ 0 \ \ \Rightarrow \ \ \left(u \ + \ \frac{B}{2} \right)^2 \ \ = \ \ \frac{B^2}{4} \ - \ C \ \ , $$
which tells us that there are no real zeroes for $ \ u \ $ if $ \ \frac{B^2}{4}  -  C \ < 0 \ \Rightarrow \ B^2 \ <  \ 4C \ \ , $ one if $ \ B^2 \ =  \ 4C \ \ , $ and two if $ \ B^2 \ > \ 4C \ \ . $  This is equivalently to saying that a parabola with its vertex at $ \ x \ = \ h \ = \ -\frac{B}{2} \ , \ y \ = \ k \ = \ C \ - \ \frac{B^2}{4} \ $ does not intersect the $ \ x-$axis in this first case, is just tangent to it in the second, and has two $ \ x-$intercepts in the third case.
With the biquadratic  $ \ x^4 \ + \ Bx^2 \ + \ C \ \ , $ which substitutes $ \ u = x^2 \ \ , $ now requires $ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \ > \ 0 \ $ for $ \ B > 0 \ $ and the polynomial cannot have a value less than $ \   \left(  \frac{B}{2} \right)^2   + \frac{B^2}{4}   -   C \ = \ C \ .  $  From this, we see that the biquadratic has  no real zeroes  for $ \ \mathbf{B > 0 \ , \ C > 0} \ \ $ and two real zeroes  for $ \ \mathbf{B > 0 \ , \ C < 0} \ \ . $  In terms of factorization, the first of these corresponds to
$$  (x^2 + \alpha^2)·(x^2 + \beta^2) \ \ = \ \  x^4 \ + \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ , $$
while the second is one case for
$$  (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \  x^4 \ + \ (\alpha^2 - \beta^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $$
with $ \ |\alpha| > |\beta| \ \ , \ \ \alpha  \ $ being imaginary.
Coming to the cases pertinent to your equations, we have $ \ B < 0 \ \ , $ which permits $ \ \left(x^2 \ + \ \frac{B}{2} \right)^2 \  \ge  \ 0 \ \ . $  The significance of this change is that the biquadratic curve has just one "turning point" for $ \ B > 0 \ \ , $ but three for $ \ B < 0 \ \ . $  The function  $ \ f(x) \ = \  x^4 \ + \ Bx^2 \ + \ C \ \   $ has even symmetry and can be thought of as "fusing" the portion of the parabola $ \ y \ = \ x^2 \ + \ Bx  \ + \ C \ $ for $ \ x \ge 0 \ $ with its "reflection" about the $ \ y-$axis.  For $ \ B > 0 \ \ , $ the vertex of the parabola is at $ \ x \ = \ -\frac{|B|}{2} \ < \ 0 \ \ , $ so the biquadratic curve only has a turning point at its $ \ y-$intercept $ \ (0 \ , \ C ) \ \ . $  However, for $ \ B < 0 \ \ , $ the vertex lies at $ \ x \ = \ \frac{|B|}{2} \ > \ 0 \ \ , $ leading to the  turning point at $ \ (0 \ , \ C ) \ \   $ and two others at $ \ x^2 \ = \  \frac{|B|}{2} \ \Rightarrow \ x \ = \ \pm \sqrt{ \frac{|B|}{2}} \ \ . $  The $ \ y-$coordinate for these points is then
$$ \left(\frac{|B|}{2} \right)^2 \ + \ B·\left(\frac{|B|}{2} \right)  \ + \ C \ \ = \ \  \frac{ B^2 }{4}   \ - \ |B|·\left(\frac{|B|}{2} \right)  \ + \ C $$ $$ = \ \   \frac{ B^2 }{4}   \ - \   \frac{ B^2}{2}    \ + \ C \ \ = \ \ C \ - \ \frac{ B^2 }{4} \ \ . $$
[The foregoing can be found more directly by using calculus.]
With this established, we can now sort  the cases for $ \ B < 0 \ \ : $
•  with the "high" turning point "below" the $ \ x-$axis $ \ [  \ \mathbf{C < 0} \  ] \ \ , $ the biquadratic curve has two $ \ x-$intercepts [the factorization is $  (x^2 + \alpha^2)·(x^2 - \beta^2) \ \ = \ \  x^4 \ - \ (\beta^2 - \alpha^2)·x^2 \ - \ \alpha^2·\beta^2 \ \ , $
with $ \ |\alpha| < |\beta| \ \ , \ \ \alpha  \ $  imaginary]
;
•  with the other turning points "above" the $ \ x-$axis, for which $ \ C \ - \ \frac{ B^2 }{4} \ > \ 0 \ \Rightarrow \ \mathbf{B^2 \ < \ 4C} \ \ , $ the curve does not intersect the $ \ x-$axis;
•  for $ \  C \ - \ \frac{ B^2 }{4} \ < \ 0 \ < \ C \ \Rightarrow \ \mathbf{0 \ < \ 4C \ < \ B^2}  \ \ , $ there are four $ \ x-$intercepts "surrounding" the three turning points [the corresponding factorization is  $ \ (x^2 - \alpha^2)·(x^2 - \beta^2) \ \ = \ \  x^4 \ - \ (\alpha^2 + \beta^2)·x^2 \ + \ \alpha^2·\beta^2 \ \ . \ ] $
We see then that for your equation (1), $ \ B = -5 \ , \ C = -36 \  $ tells us that the equation has two real zeroes; its factorization is $ \ (x^2 - 9)·(x^2 + 4) \  \ . $  Equation (2) has $ \ B = -13 \ , \ C = 36 \ $ $\Rightarrow \ B^2 = 169 \ > \ 4C = 144 \ > \ 0 \ \ , $ hence there are four real zeroes; the expression factors as $ \ (x^2 - 9)·(x^2 - 4) \  \ . $  Finally, with equation (3), we "factor out" a $ \ 4 \ \   $ (which has no effect on the solutions) to obtain $ \ B = -\frac52 \ , \ C = \frac{25}{4} \ \Rightarrow \ B^2 = \frac{25}{4} \ < \ 4C = 25 \ \ , $ indicating that there are no real zeroes.  (The factorization here is not very tidy.)
A table of the conditions for the number of real zeroes is presented below.

