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If (ax+b)(bx+a)=26x^2+cx+26, where a, b, and c are distinct integers, what is the minimum possible value of c, the coefficient of x?

 Nov 6, 2018

Best Answer 

 #1
avatar+6252 
+1

(ax+b)(bx+a)=abx2+(a2+b2)x+ab=26x2+cx+26ab=26c=a2+b2

 

so basically we're minimizing a2+b2 subject to ab=26and subject to a,b,cZ

 

26 can be factored as (2,13), (1,26)22+132=17312+262=676and 173 is the smaller of those two so c=173

.
 Nov 6, 2018
 #1
avatar+6252 
+1
Best Answer

(ax+b)(bx+a)=abx2+(a2+b2)x+ab=26x2+cx+26ab=26c=a2+b2

 

so basically we're minimizing a2+b2 subject to ab=26and subject to a,b,cZ

 

26 can be factored as (2,13), (1,26)22+132=17312+262=676and 173 is the smaller of those two so c=173

Rom Nov 6, 2018

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