### Instruction: If the quadratic $x^2+4mx+m$ has exactly one real root, find the positive value of $m$.### Response: I know that a quadratic equation has exactly one real root when its discriminant is zero, so I can use the formula $b^2-4ac=0$ to find $m$.
In this case, $a=1$, $b=4m$, and $c=m$, so I get $(4m)^2-4(1)(m)=0$.
Expanding and simplifying, I get $16m^2-4m=0$.
I can factor out a common factor of $4m$, and get $4m(4m-1)=0$.
Since $m$ is positive, I can ignore the solution $m=0$ and focus on the other one: $4m-1=0$.