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The quadratic equation x^2 - ax + 2016 + 5 = 0 has two postiive integer solutions.  Find the minimum value of a.

 Jan 7, 2021
 #1
avatar+114361 
+1

x^2  - ax  +  2016 +  5  =  0

 

factors of 2021 =   1, 43, 47, 2021

 

x^2   - ax  + 2021  =   0       factor as

 

(x  - 43) ( x  - 47)  =  0

 

Min  value  of a  = -(43 + 47)  =  -90

 

 

cool cool cool

 Jan 7, 2021
edited by CPhill  Jan 8, 2021
 #2
avatar+428 
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We might also take the discriminant, b^2 - 4ac.

 

We have that (-a)^2 - 4 * 2016 * 1 > 1, so a^2 - 8064 > 0. Thus, a^2 > 8064, and a > ±√8064, so a> about 89.8 and a< about -89.8. However, we would like to find the integer solutions, which means a must be an integer, which means that a ≤ -90.

 

(By the way CPhill, you have a typo on the first line)

 Jan 8, 2021

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