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# Algebra

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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.

Jul 25, 2024

#1
+1790
+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! :)

Jul 25, 2024
edited by NotThatSmart  Jul 25, 2024

#1
+1790
+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
edited by NotThatSmart  Jul 25, 2024