The roots of the equation 1/a+b+x = 1/a + 1/b + 1/x are a. : Quant Question Archive [LOCKED]
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# The roots of the equation 1/a+b+x = 1/a + 1/b + 1/x are a.

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The roots of the equation 1/a+b+x = 1/a + 1/b + 1/x are a. [#permalink]

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29 Jan 2004, 14:21
This topic is locked. If you want to discuss this question please re-post it in the respective forum.

The roots of the equation 1/a+b+x = 1/a + 1/b + 1/x are

a. a, b
b. a+b, a-b
c. -a, -b
d. 1/a+b, 1/a-b
e. Beats me!
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29 Jan 2004, 15:05
Oh God , is there a way to keep myself from getting distracted.

Anyway is it C) -a, -b
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29 Jan 2004, 15:27

How did you work this one old chap?
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Pls include reasoning along with all answer posts.
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29 Jan 2004, 15:37
Hi sunniboy, can I meet ur friends . Took me around 2 mins but I guess its the only way. Cross multiply, then u can cancel abx and then derive at the equation in the form of ax^2+bx+c=0, we get the equation (a+b)x^2 + (a+b)^2 x + ab(a+b), I think now u will be able to find out the roots.
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30 Jan 2004, 08:27
It might actually be simpler in this case to check one of each pair of roots - if one of them doesn't work - the answer choice is definitely wrong.

A) 1/a+b+a is not equal to 1/a + 1/b + 1/a - this is the concept that says that one cannot distribute a fraction under a common numerator...

B) for the same reason - 1/a+b+a+b is not 1/a + 1/b + 1/a+b for any a and b.

C) BUT - 1/a+b-a is just 1/b. 1/a + 1/b - 1/a is exactly the same. Ditto with -b. and you have your solution.

Lofty algebra is not always the right way to go - if it gets messy - look for an alternative....
roots...   [#permalink] 30 Jan 2004, 08:27
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