+1493 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.
+1946 First, let's move all terms to one side and combine all like terms. We get
x2+(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>417−4>mm<13
From the second interval, we have
17−m<−421<mm>21
Thus, we have two intervals for m. So our final answer, converting in interval notation, is
m=(−∞,13)U(21,∞)
Thanks! :)
