What is the largest value of $x$, if $\frac{x}{5} + \frac{1}{5x} = \frac{1}{2}$?
We multiply both sides of the equation by $10x$ to clear the fractions, leaving us with $2x^2 + 2 = 5x$. Rearranging the terms, we have $2x^2 - 5x + 2 = 0$. We can now solve for $x$ by factoring: $(2x - 1)(x - 2) = 0$. We could also use the quadratic formula:  $$x = \frac{5 \pm \sqrt{(-5)^2 - 4(2)(2)}}{4}.$$Either way, we find that $x = 1/2$ or $x = 2$. Since we want the largest value of $x$, our answer is $\boxed 2$.
2
Level 3
Algebra
Problem: What is the largest value of $x$, if $\frac{x}{5} + \frac{1}{5x} = \frac{1}{2}$?
Solution:<|START_OF_LATENT|><|PRIOR_PREFIX|>To solve the equation $\frac{x}{5} + \frac{1}{5x} = \frac{1}{2}$, first eliminate the fractions by multiplying through by $10x$, the least common multiple of the denominators. This gives:

$10x \cdot \frac{x}{5} + 10x \cdot \frac{1}{5x} = 10x \cdot \frac{1}{2}$.

Simplifying each term results in:

$2x^2 + 2 = 5x$.

Rearranging this equation leads to:

$2x^2 - 5x + 2 = 0$.

Next, apply the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$, where $a = 2$, $b = -5$, and $c = 2$. Calculate the discriminant:

$b^2 - 4ac = (-5)^2 - 4 \cdot 2 \cdot 2 = 25 - 16 = 9$.

Since the discriminant is positive, there are two real solutions. Now, substitute into the quadratic formula:

$x = \frac{5 \pm \sqrt{9}}{2 \cdot 2} = \frac{5 \pm 3}{4}$.

This results in two potential solutions:

1. $x = \frac{8}{4} = 2$.
2. $x = \frac{2}{4} = \frac{1}{2}$.

To find the largest value of $x$, compare the two solutions. The larger solution is $2$. Thus, the final answer is $2$.<|END_OF_LATENT|> The final answer is $2$.