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

 

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

 

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

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 Nov 6, 2018
 #1
avatar+4459 
+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

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