+608 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 First, let's simplify the equation so that all terms are on one side.
Combining all like terms and moving them all to one side, we get the equation
x2+(17−m)x+4=0
Now, since it has two distinct roots, the descriminant of the quadratic must be greater than 0.
Thus, we have the equation
(17−m)2−4(1)(4)>0(17−m)2>16
We get two roots from this equation.
Let's calculate both of them. For the first one, we have
17−m>417−4>mm<13
For the second root, we have
17−m<−421<mm>21
Thus, in interval notation, we have m=(−∞,13)U(21,∞)
Thanks! :)
+1950 First, let's simplify the equation so that all terms are on one side.
Combining all like terms and moving them all to one side, we get the equation
x2+(17−m)x+4=0
Now, since it has two distinct roots, the descriminant of the quadratic must be greater than 0.
Thus, we have the equation
(17−m)2−4(1)(4)>0(17−m)2>16
We get two roots from this equation.
Let's calculate both of them. For the first one, we have
17−m>417−4>mm<13
For the second root, we have
17−m<−421<mm>21
Thus, in interval notation, we have m=(−∞,13)U(21,∞)
Thanks! :)
