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January 29, 2015

January 29, 2015

Posted by **Kiara** on Wednesday, July 9, 2014 at 4:03pm.

p5 + 2x2 + 3px + 7 + 8 = -2:

- Algebra II - Bogus -
**Steve**, Wednesday, July 9, 2014 at 4:06pmI suspect typos. The function is littered with junk.

p?

+7+8?

- Algebra II -
**Reiny**, Wednesday, July 9, 2014 at 4:14pmI will assume p5 is meant to be p^5

and 2x2 is meant to be 2x^2 , then

p^5 + 2x^2 + 3px + 17 = 0

2x^2 + 3px + p^5 + 17 = 0

a quadratic with a = 2 , b = 3p and c = p^5 + 17

for real roots:

9p^2 - 4(2)(p^5 + 17) ≥ 0

9p^2 - 8p^5 - 136 ≥ 0

using Wolfram to solve, I got p < -1.69

http://www.wolframalpha.com/input/?i=9p%5E2+-+8p%5E5+-+136+%3D+0

check your typing, as long as p < -1.69 you will have 2 real solutions each time

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