+123 What is the smallest positive value of m so the equation 10x^2-mx+420=0 has integral solutions?
+129852 10x^2 -mx - 420 = 0 factor out the 10
10 (x^2 - (1/10)m * x - 42) = 0 divide both sides by 10
x^2 - (m/10) x - 42 = 0
We need two integers that multiply to - 42 and whose sum is an integer for (m/10)
Note that the smallest positive value for m that produces an integer for m/10 is when m = 10
So
x^2 - (1/10) (10)x - 2 = 0
x^2 - x - 42 = 0
(x - 7) ( x + 6) = 0
x = 7 and x = -6
So...m = 10 is the smallest value for m that prduces integer solutions for 10x^2 - mx - 420 = 0
+123 Let $p$ and $q$ be the solutions the the equation $10x^2 - mx + 420 = 0$. We use the fact that the sum and product of the roots of a quadratic equation $ax^2+bx+c = 0$ are given by $-b/a$ and $c/a$, respectively, so $p+q = m/10$ and $pq = 420/10 = 42$. Since $m = 10(p+q)$, we minimize $m$ by minimizing the sum $p+q$. Since $p$ and $q$ are integers and multiply to 42, the possible values of $(p,q)$ are $(1,42),(2,21),(3,14),(6,7),(7,6),(14,3),(21,2),(42,1)$. (Note that if $p$ and $q$ are both negative, then $p+q$ is negative, so $m$ would be negative, which is excluded by the problem.) The sum $p+q$ is minimized when $(p,q) = (6,7)$ or $(7,6)$. In either case, $m = 10(p+q) = 10(6+7) =130
