Finding smallest value for parameterised answer that satisfies condition - matlab

Question

I want to find the smallest integer l that satisfies l^2 >= x, and mod(l,2)=0.
In the following example x=75, and hence l=10, since the previous even number doesn't fulfil the inequation: 8^2 <= 75 <= 10^2
I have tried this (ignoring the even-number requirement, which I can't to work):
syms l integer
eqn1 = l^2 >= 75;
% eqn2 = mod(l,2) == 0;
[sol_l, param, cond] = solve(eqn1, l, 'ReturnConditions', true);
But this does not give me anything helpful directly:
sol_l =
k
param =
k
cond =
(75^(1/2) <= k | k <= -75^(1/2)) & in(k, 'integer')
I would like to evaluate the conditions on the parameter and find the smallest value that satisfies the conditions.
Also, I would like to enforce the mod(l,2)=0 condition somehow, but I don't seem to get that work.

Answer 1

Using the solve for this task is like using a cannon to kill a mosquito. Actually, the answer of Lidia Parrilla is good and fast, although it can be simplified as follows:
l = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l = l + 1;
end
% if x = 75, then l = 10
But I would like to point out something that no one else noticed. The condition provided by the solve function for l^2 >= 75 is:
75^(1/2) <= k | k <= -75^(1/2)
and it's absolutely correct. Since l is being raised to the power of 2, and since a negative number raised to the power of 2 produces a positive number, the equation will always have two distinct solutions: a negative one and a positive one.
For x = 75, the solutions will be l = 10 and l = -10. So, if you want to find the smallest number (and a negative number is always smaller than a positive one), the right solution will be:
l = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l = l + 1;
end
l = l * -1;
If you want to return both solutions, the result will be:
l_pos = ceil(sqrt(x));
if (mod(x,2) ~= 0)
l_pos = l_pos + 1;
end
l_neg = l_pos * -1;
l = [l_neg l_pos];

Answer 2

I guess the easiest solution without employing the inequality and solve function would be to find the exact solution to your equation l^2 >= x, and then finding the next even integer. The code would look like this:
x = 75;
y = ceil(sqrt(x)); %Ceil finds the next bigger integer
if(~mod(y,2)) %If it's even, we got the solution
sol = y;
else %If not, get the next integer
sol = y+1;
end
The previous code gives the correct solution to the provided example (x = 75; sol = 10)