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Replies: 9   Last Post: Oct 18, 1999 10:23 PM

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 dannyboy@here.com Posts: 682 Registered: 12/6/04
Posted: Oct 17, 1999 5:50 PM

On Sun, 17 Oct 1999 15:06:40 -0400, "Nick Marques" <nickjm@jnlk.com>
wrote:

>If I have some nasty numbers in quadratic form, how do I factor and solve
>
>-4.9x^2+5.5x-105=0
>That factors to x(-4.9x+5.5) = -105

Gets you nowhere. You are looking for two factors of the form
(Ax+B)(Cx+D). The + might be -. However, MUCH easier to use the
quadratic formula, especially in this case as you will see.

4.9x^2 - 5.5x + 105 = 0

Use the formula:

x = (-b +-sqrt(b^2-4ac)/2a to get

x = (5.5 +-sqrt((5.5)^2 - 4*4.9*105))/(2*4.9)

Out with the calculator, and it looks like they might be complex
roots.

Dan.

Date Subject Author
10/17/99 Nick Marques
10/17/99 Wilson Figueroa
10/17/99 Physics Boy
10/17/99 dannyboy@here.com
10/18/99 Tom Davidson
10/18/99 Richard Carr
10/18/99 Tom Davidson
10/18/99 dannyboy@here.com
10/18/99 Tom Davidson
10/18/99 Kyle R. Hofmann