+1855 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.
+1950 We can write an inequality to solve this problem. We have
x2+(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>417−4>mm<13
solving the other root, we get
17−m<−421<mm>21
Thus, in interval notation, we have
(−∞,13)U(21,∞)
So (−∞,13)U(21,∞) is our final answer.
Thanks! :)
+1950 We can write an inequality to solve this problem. We have
x2+(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>417−4>mm<13
solving the other root, we get
17−m<−421<mm>21
Thus, in interval notation, we have
(−∞,13)U(21,∞)
So (−∞,13)U(21,∞) is our final answer.
Thanks! :)
