Posted by Kiara on Wednesday, July 9, 2014 at 4:03pm.
Explain, without doing any calculations, why there cannot be a Real solution to the equation
p5 + 2x2 + 3px + 7 + 8 = 2:

Algebra II  Bogus  Steve, Wednesday, July 9, 2014 at 4:06pm
I suspect typos. The function is littered with junk.
p?
+7+8?

Algebra II  Reiny, Wednesday, July 9, 2014 at 4:14pm
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
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