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.

onyuIee Jul 25, 2024

#1**+1 **

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! :)

NotThatSmart Jul 25, 2024

#1**+1 **

Best Answer

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! :)

NotThatSmart Jul 25, 2024