Posted by maddie on Sunday, May 24, 2009 at 11:36am.
I'm quite clueless as to how to solve this. I know the quadratic equation but am not sure how to get this equation into the format to plug in to the quadratic...
Solve the quadratic equation using the formula [1/(1+x)][1/(3x)]=(6/35)
Show your work

algebra II  bobpursley, Sunday, May 24, 2009 at 11:46am
multiply both sides by 35(x+1)(x3)
then
35(x3)+35(x+1)=6(x3)(x+1)
and you should be able to assemble it from there.

algebra II  maddie, Sunday, May 24, 2009 at 11:50am
so should i just multiply that all out and then plug it in?

algebra II  drwls, Sunday, May 24, 2009 at 11:51am
Write the two terms on the left with a common denomimnator. Then convert to standard quadratic form.
[(3x)  (1+x)]/[(1+x)(3x)] = 6/35
(22x)/[(1+x)(3x)] = 6/35
(1x) = (3/35)[(1+x)(3x)]
35 35x = 3[x^2 +2x +3)
3x^2 41x +26 = 0
(3x 2)(x13)
Check my work. No guarantees

algebra II  drwls, Sunday, May 24, 2009 at 11:52am
We were lucky that one factored fairly easily. x = 2/3 or 13

algebra II  maddie, Sunday, May 24, 2009 at 12:26pm
i got the same answer but used the quadratic :) thank you for all your help everyone!
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