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.

Rangcr897 Jun 8, 2024

#1**+1 **

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

We get

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

Let's note that if roots are real, then the descriminant must be greater or equal to 0.

We have

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

From here, there are 2 ways to go. First, we have

\(17 - m \geq 4 \\ 17 - 4 \geq m \\ m \leq 13 \)

and, we have

\( 17 - m \leq -4 \\ 21 \leq m \\ m \geq 21\)

This means our final answer is

\(m = [-\infty , 13]\bigcup [21, \infty]\)

Thanks! :)

NotThatSmart Jun 9, 2024