## cos00155079 Group Title p^2x^2+2px+1=0 solve using the quadratic formula 2 years ago 2 years ago

1. cos00155079 Group Title

|dw:1335056613189:dw| what is a? or how do you solve for a here?

2. bmp Group Title

Assuming p is a real number, p^2 = a, 2p = b and c = 1, because they are all constants.

3. cos00155079 Group Title

so what is the rest...? do i just ignore it? the xs?

4. bmp Group Title

x is the x in the quadratic equation. It will become:$x = (-2p \pm \sqrt{(2p)^{2} - 4p^{2}})/(2p^{2})$

5. imagreencat Group Title

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). :)

6. exraven Group Title

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}$

7. bmp Group Title

@imagreencat Indeed, that's easier to solve :-)

8. cos00155079 Group Title

thanks everyone :)

9. imagreencat Group Title

You're welcome!