+303 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.
+1790 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
\(x^2 + (17 - m)x + 4 = 0 \)
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
\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)
For the second root, we have
\( 17 - m < -4 \\ 21 < m \\ m > 21\)
Thus, in interval notation, we have \(m = ( -\infty, 13) U ( 21, \infty) \)
Thanks! :)
+1790 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
\(x^2 + (17 - m)x + 4 = 0 \)
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
\(17 - m > 4 \\ 17 - 4 > m \\ m < 13 \)
For the second root, we have
\( 17 - m < -4 \\ 21 < m \\ m > 21\)
Thus, in interval notation, we have \(m = ( -\infty, 13) U ( 21, \infty) \)
Thanks! :)
