+0  
 
0
30
1
avatar

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?

Guest Nov 6, 2018

Best Answer 

 #1
avatar+2786 
+1

\((a x + b)(b x + a) = ab x^2 + (a^2+b^2)x+ab = 26x^2 + c x + 26\\ \\ ab = 26\\ c = a^2 + b^2\)

 

\(\text{so basically we're minimizing }a^2+b^2 \text{ subject to }ab=26\\ \text{and subject to }a,b,c \in \mathbb{Z}\)

 

\(26 \text{ can be factored as }(2,13),~(1,26)\\ 2^2 + 13^2 = 173\\ 1^2 + 26^2 = 676\\ \text{and 173 is the smaller of those two so }\\ c=173\)

Rom  Nov 6, 2018
 #1
avatar+2786 
+1
Best Answer

\((a x + b)(b x + a) = ab x^2 + (a^2+b^2)x+ab = 26x^2 + c x + 26\\ \\ ab = 26\\ c = a^2 + b^2\)

 

\(\text{so basically we're minimizing }a^2+b^2 \text{ subject to }ab=26\\ \text{and subject to }a,b,c \in \mathbb{Z}\)

 

\(26 \text{ can be factored as }(2,13),~(1,26)\\ 2^2 + 13^2 = 173\\ 1^2 + 26^2 = 676\\ \text{and 173 is the smaller of those two so }\\ c=173\)

Rom  Nov 6, 2018

37 Online Users

avatar
avatar
avatar

New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.