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The quadratic equation $x^2-mx+24 = 10$ has roots $x_1$ and $x_2$. If $x_1$ and $x_2$ are integers, how many different values of $m$ are possible?

 Jun 8, 2024
 #1
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First, let's subtract 10 from both sides. \(x^2-mx+14\)

The answer to this question lies in the number of possible multiples 14 has. 

 

For every mutliple, there are 4 possibles. 

Say the two multiples are \(a, b\)

We have \((x-a)(x-b), (x+a)(x+b), (x+a)(x-b), (x-a)(x+b)\)

 

For example, 2 and 7. 

\((x-7)(x+2), (x-2)(x+7), (x-2)(x-7), (x+7)(x+2)\)These all give different values for m. 

 

We have only 2 possible groups. 1 and 14, 2 and 7. 

 

This means there are 4*2=8 different values for m

 

Thanks! :)

 Jun 8, 2024

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