I assume x is a real variable. Note that √(25x²) is a non-negative number for any real x.
Meanwhile, we observe that for -2x √(100x),
• it's only defined for x ≥ 0, and
• if x ≥ 0, then multiplying √(100x) by -2x makes the overall product non-positive, i.e. -2x √(100x) ≤ 0
In short, this eliminates any non-zero real solution, and we're left with x = 0.
We can arrive at the same result by working with the equation:
√(25x²) + 2x √(100x) = 0
5|x| + 20 x√x = 0
since √(x²) = |x| for all x.
Recall that |x| = x if x ≥ 0, and |x| = -x if x < 0.
• If x ≥ 0, the equation reduces to
5x + 20 x√x = 0
5x (1 + 4√x) = 0
5x = 0 or 1 + 4√x = 0
x = 0 or √x = -1/4
But √x ≥ 0 for x ≥ 0, so we throw out the second solution.
• If x < 0, the equation instead reduces to
-5x + 20 x√x = 0
-5x (1 - 4√x) = 0
-5x = 0 or 1 - 4√x = 0
x = 0 or √x = 1/4
x = 0 or x = 1/16
But we assumed x is negative, so we throw out both choices and get no solution for this case.