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What is the smallest positive value of m so the equation 10x^2-mx+420=0 has integral solutions?

 Mar 9, 2019
 #1
avatar+111395 
0

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

 

 

cool cool cool

 Mar 9, 2019
 #2
avatar+118 
+1

It is wrong. How can you factor out the 10 when m is a variable?

IneedHALP  Mar 11, 2019
 #3
avatar+111395 
0

Check for yourself, here, that  any integer  m  between 1  and 9  inclusive will not provide integer solutions

 

https://www.wolframalpha.com/input/?i=10x%5E2+-+10x+-420+%3D+0

 

The smallest positive integer m that gives integer solutions is when m = 10

 

cool cool cool

CPhill  Mar 11, 2019
 #4
avatar+118 
0

The system that I put the answer into says it is wrong.

IneedHALP  Mar 11, 2019
 #5
avatar+118 
+1

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

 Mar 16, 2019

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