+782 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.
+1365 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! :)
