What is the sum of the non-integral solutions of the equation

\( \left| \begin{array}{ccc} x & 3 & 4 \\ 5 & x & 5 \\ 4 & 2 & x \end{array} \right|=0\hspace{2mm}?\)

What is the sum of the non-integral solutions of the equation

\( \left| \begin{array}{ccc} x & 3 & 4 \\ 5 & x & 5 \\ 4 & 2 & x \end{array} \right|=0\hspace{2mm}?\)

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TopNewestReduce to a \(2 \times 2\) matrix: \(\left[ \begin{array}{cc} 10-4x & x^2-10 \\ 12-2x & 8-3x \end{array} \right]= 0\)

Then to a \(1 \times 1\) matrix: \((8-3x)(10-4x)-(12-2x)(x^2-10) = 0\)

Which then becomes \(x^3 - 41x + 100 = 0\)

Then after trial and error, x = 4 is one of the non-zero solutions – Hero Miles · 3 years, 11 months ago

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– Krishna Jha · 3 years, 11 months ago

Please see the edit now..And please could you give a solution other than hit and trial??Like using synthetic division??Log in to reply

– Hero Miles · 3 years, 11 months ago

This solution was created using the Carroll-Dodson Condensation Method. Synthetic division in this case has no meaningful purpose. However, I will see what I can do as far as posting a better solution.Log in to reply

– Mark Hennings · 3 years, 11 months ago

Why not just calculate the \(3\times3\) determinant directly? The C-D method requires the calculation of five \(2\times2\) determinants, when only three are necessary: \[ 0 \; = \; x(x^2-10) - 3(5x-20) + 4(10-4x) \; = \; x^3 - 41x + 100 \] About the only short-cut to looking for the integer root of this cubic is to remember that any integer root \(n\) has to be a factor of \(100\). Testing \(\pm1,\pm2,\pm4,\pm5,\ldots\) we find \(n=4\) quite quickly. We could trim the testing a little by observing that \(|x^3-41x| \le 8 + 82 = 90 < 100\) for \(|x| \le 2\), and hence \(|n| \ge 4\). Thus we can start testing with \(n=\pm4\), and hit paydirt first time. We might not have been so lucky, and it isn't really necessary. Testing on a predetermined finite set (the factors of \(100\)) is not really trial and error.Log in to reply