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# Algebra

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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?

Jul 23, 2024

#1
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First, let's simplify the equation so that we have a quadratic equal to 0. We have

\( x^2 – mx + 24 = 10    \\ x^2 – mx + 14 = 0  \)

The coefficient of x is the sum of the factors of 14

The integer factors of 14 are (2, 7) and (–2, –7)

and (1, 14) and (–1, –14)

So m could equal 9 or –9 or 15 or –15

Therefore, m could have 4 possible values

Thanks! :)

Jul 23, 2024
edited by NotThatSmart  Jul 23, 2024

#1
+1661
+1

First, let's simplify the equation so that we have a quadratic equal to 0. We have

\( x^2 – mx + 24 = 10    \\ x^2 – mx + 14 = 0  \)

The coefficient of x is the sum of the factors of 14

The integer factors of 14 are (2, 7) and (–2, –7)

and (1, 14) and (–1, –14)

So m could equal 9 or –9 or 15 or –15

Therefore, m could have 4 possible values

Thanks! :)

NotThatSmart Jul 23, 2024
edited by NotThatSmart  Jul 23, 2024