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.

learnmgcat Jul 14, 2024

#1**+1 **

First, let's move all terms to one side and combine all like terms. We get

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

If there are two distict real roots, then the descriminant must be greater than 0.

Thus, we have

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

We must take both intervals for m.

Taking the first interval, we have

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

From the second interval, we have

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

Thus, we have two intervals for m. So our final answer, converting in interval notation, is

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

Thanks! :)

NotThatSmart Jul 15, 2024