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# HELP! EXTREMELY HARD PROBLEM!

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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
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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   Mar 9, 2019
#2
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It is wrong. How can you factor out the 10 when m is a variable?

IneedHALP  Mar 11, 2019
#3
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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   CPhill  Mar 11, 2019
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The system that I put the answer into says it is wrong.

IneedHALP  Mar 11, 2019
#5
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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