I suspect typos. The function is littered with junk.
p?
+7+8?
Explain, without doing any calculations, why there cannot be a Real solution to the equation
p5 + 2x2 + 3px + 7 + 8 = -2:
2 answers
I 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
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
2 answers