+1262 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.
+953 We can write an inequality to solve this problem. We have
\(x^2 + (17 - m)x + 4 > 0\)
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
\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)
solving the other root, we get
\( 17 - m < -4 \\ 21 < m \\ m > 21 \)
Thus, in interval notation, we have
\(( -\infty , 13)U ( 21, \infty)\)
So \(( -\infty , 13)U ( 21, \infty)\) is our final answer.
Thanks! :)
+953 We can write an inequality to solve this problem. We have
\(x^2 + (17 - m)x + 4 > 0\)
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
\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)
solving the other root, we get
\( 17 - m < -4 \\ 21 < m \\ m > 21 \)
Thus, in interval notation, we have
\(( -\infty , 13)U ( 21, \infty)\)
So \(( -\infty , 13)U ( 21, \infty)\) is our final answer.
Thanks! :)
