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Re: x^2 + x - 40 = 0 [#permalink]
2
IlCreatore wrote:
The solutions to the equations are \(\frac{-1\pm \sqrt161}{2}\), then given that \(sqrt161>12\), the two roots are both positive numbers. Thus, when they are summed to 1 and taken absolute value they keep being positive and greater than 5 in both the occasions.

Thus answer is A

Another good explanation,
Giving you 1 more kudos.
It was a good question.
Yes, in both scenarios, Quantity A always greater.
So Choice A.
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Re: x^2 + x - 40 = 0 [#permalink]
would youplz give anoyher solution?
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Re: x^2 + x - 40 = 0 [#permalink]
Hello from the GRE Prep Club BumpBot!

Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).

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Re: x^2 + x - 40 = 0 [#permalink]
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