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# Algebra

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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

#1
+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
+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! :)

NotThatSmart Jun 27, 2024