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Let $m$ be a real number. If the quadratic equation $x^2+mx+4 = 2x^2 + 17x + 8$ has two distinct real roots, then what are the possible values of $m$? Express your answer in interval notation.

 Jun 27, 2024

Best Answer 

 #1
avatar+1230 
+1

We can write an inequality to solve this problem. We have

\(x^2 + (17 - m)x + 4 > 0\)

 

The rule states that if there are two distinct roots, then the descriminant must be greater thnan  0. Thus, we have the equation

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16\)

 

We take both roots from the equation and solve both. We have

\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)

 

solving the other root, we get

\( 17 - m < -4 \\ 21 < m \\ m > 21 \)

 

Thus, in interval notation, we have

\(( -\infty , 13)U ( 21, \infty)\)

 

So \(( -\infty , 13)U ( 21, \infty)\) is our final answer. 

 

Thanks! :)

 Jun 27, 2024
 #1
avatar+1230 
+1
Best Answer

We can write an inequality to solve this problem. We have

\(x^2 + (17 - m)x + 4 > 0\)

 

The rule states that if there are two distinct roots, then the descriminant must be greater thnan  0. Thus, we have the equation

\((17 - m)^2 - 4(1)(4) > 0 \\ (17 - m)^2 > 16\)

 

We take both roots from the equation and solve both. We have

\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)

 

solving the other root, we get

\( 17 - m < -4 \\ 21 < m \\ m > 21 \)

 

Thus, in interval notation, we have

\(( -\infty , 13)U ( 21, \infty)\)

 

So \(( -\infty , 13)U ( 21, \infty)\) is our final answer. 

 

Thanks! :)

NotThatSmart Jun 27, 2024

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