The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.

thanks

2. Originally Posted by gracey
The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.

thanks
Possible values based on what criteria? As it stands p could have any value.

-Dan

3. Originally Posted by gracey
The quadratic equation 2x^2 + (p+1)x + 8 = 0 find possible values of p.

thanks

$x=\frac{-(p+1) \pm \sqrt{(p+1)^2 - 4(8)(2)}}{4}$

$p+1$ is restricted by the radical or discriminant in order for x to have real solutions.

$(p+1)^2 - 4(8)(2) \ge 0$

$p^2 + 2p -63 \ge 0$

$(p+9)(p-7) = 0$

$p = -9, 7$

You have to test the three values that are less than -9, between -9 and 7, and greater than 7 to see which sub domains hold for the inequality.

It turns out that $p \le 9, p \ge 7$ OR $p: (-\infty, -9] \cap [7, \infty)$

4. Ah but the given condition is apparently nothing more than that the equation be valid. No one said we can't have complex zeros. I was waiting for verification on that point.

-Dan

5. Originally Posted by topsquark
Ah but the given condition is apparently nothing more than that the equation be valid. No one said we can't have complex zeros. I was waiting for verification on that point.

-Dan
True, but when is a problem stated with all the given conditions. A lot is assumed in elementary problems, let alone all mathematical problems outside of a full proof.

6. thankyou everyone