Here's the question you clicked on:
cos00155079
p^2x^2+2px+1=0 solve using the quadratic formula
|dw:1335056613189:dw| what is a? or how do you solve for a here?
Assuming p is a real number, p^2 = a, 2p = b and c = 1, because they are all constants.
so what is the rest...? do i just ignore it? the xs?
x is the x in the quadratic equation. It will become:\[x = (-2p \pm \sqrt{(2p)^{2} - 4p^{2}})/(2p^{2})\]
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). :)
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\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]
@imagreencat Indeed, that's easier to solve :-)