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cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
dw:1335056613189:dw what is a? or how do you solve for a here?
 2 years ago

bmp Group TitleBest ResponseYou've already chosen the best response.0
Assuming p is a real number, p^2 = a, 2p = b and c = 1, because they are all constants.
 2 years ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
so what is the rest...? do i just ignore it? the xs?
 2 years ago

bmp Group TitleBest ResponseYou've already chosen the best response.0
x is the x in the quadratic equation. It will become:\[x = (2p \pm \sqrt{(2p)^{2}  4p^{2}})/(2p^{2})\]
 2 years ago

imagreencat Group TitleBest ResponseYou've already chosen the best response.2
Actually, it's just (px)^2 + 2px + 1 = 0. We can just easily factor this especially since C here only has (1)(1) as its factors. (px + 1)(px + 1). :)
 2 years ago

exraven Group TitleBest ResponseYou've already chosen the best response.0
a quadratic equation can be written in the form:\[ax^2 + bx + c = 0\]then we get\[a = p^2\]\[b = 2p\]\[c = 1\]after that you just need to plug them to the quadratic formula\[x = \frac{b \pm \sqrt{b^2  4ac}}{2a}\]
 2 years ago

bmp Group TitleBest ResponseYou've already chosen the best response.0
@imagreencat Indeed, that's easier to solve :)
 2 years ago

cos00155079 Group TitleBest ResponseYou've already chosen the best response.0
thanks everyone :)
 2 years ago

imagreencat Group TitleBest ResponseYou've already chosen the best response.2
You're welcome!
 2 years ago
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