$\displaystyle \frac{90}{s+5} + \frac{90}{s-5} = 2\frac{1}{8} $ If I want to enter this into quadratic formula, do I need to change it someway? The coefficients are in fractions and that's what confuses me.
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Hi reallylongnicknam. put the equation into this form a^2 +bx +C = 0 bjh
Firstly, multiple both sides by (s+5)(s-5). What do you get?
$\displaystyle s^2 -25 $ I'm thinking there is cross multipling with those 90s.
Last edited by reallylongnickname; Mar 30th 2011 at 06:28 PM.
I mean $\displaystyle \displaystyle \frac{90}{s+5} \times (s+5)(s-5)+ \frac{90}{s-5}\times (s+5)(s-5) = 2\frac{1}{8}\times (s+5)(s-5) $ What do you get?
$\displaystyle 90 (s-5) + 90 (s+5) = 2\frac{1}{8} (s-5)(s+5) $
Good, now expand each term and group where they are like. What do you get?
Originally Posted by reallylongnickname $\displaystyle 90 (s-5) + 90 (s+5) = 2\frac{1}{8} (s-5)(s+5) $ OK so far...continue...easier if you change 2 1/8 to 17/8
$\displaystyle -\frac{17}{8 }s^2 + 180s + 25 = 0 $ Would I divide all coefficients by 8 to remove the fraction before I can plug this into the quadratic formula?
Last edited by reallylongnickname; Mar 31st 2011 at 03:36 AM.
You don't have to: a, b, and c in "$\displaystyle ax^2+ bx+ c= 0$" don't have to be integers but you will probably find the arithmetic easier.
Originally Posted by reallylongnickname $\displaystyle -\frac{17}{8 }s^2 + 180s + 25 = 0 $ That is not correct. 90(S - 5) + 90(S + 5) = 180S 17/8(S - 5)(S + 5) = 17/8(S^2 - 25) So: (17/8)S^2 - (17/8)25 - 180S = 0 17S^2 / 8 - 180S - 425 / 8 = 0 Multiply by 8: 17S^2 - 1440S - 425 = 0 OK?
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