Find the sum of all numbers $x$ satisfying $x+25/x = 10.$
Multiplying both sides by $x$ and then subtracting $10x$ from each side gives $x^2 - 10 x + 25 = 0.$ The quadratic factors to give $(x-5)^2 = 0,$ so $x-5 = 0,$ and $x=5$ is the only solution. Thus the answer is $\boxed{5}.$

Note: We might want to use the fact that the sum of the solutions to a quadratic $ax^2+bx+c = 0$ is given by $-b/a,$ but be careful! That fact counts double roots twice for the purpose of the sum, but this problem only counts it once, since $x=5$ is the only solution.
5
Level 3
Algebra
Problem: Find the sum of all numbers $x$ satisfying $x+25/x = 10.$
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation \( x + \frac{25}{x} = 10 \), first, multiply both sides by \( x \) to eliminate the fraction, resulting in \( x^2 + 25 = 10x \). Rearranging this gives \( x^2 - 10x + 25 = 0 \). 

Next, recognize that this is a quadratic equation in standard form \( ax^2 + bx + c = 0 \) where \( a = 1 \), \( b = -10 \), and \( c = 25 \). 

To find the roots, apply the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \). Calculate the discriminant: \( b^2 - 4ac = (-10)^2 - 4 \cdot 1 \cdot 25 = 100 - 100 = 0 \). 

Since the discriminant is zero, there is exactly one real root. Substitute into the quadratic formula: \( x = \frac{10 \pm 0}{2} = \frac{10}{2} = 5 \). 

Thus, the only solution to the equation is \( x = 5 \). 

Finally, since the problem asks for the sum of all solutions and there is only one solution, the sum is simply \( 5 \).<|END_OF_LATENT|> The final answer is $5$.