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# Is A positive? 1) x^2 - 2x +A is positive for all x 2) A*x^2

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Director
Joined: 03 May 2007
Posts: 867

Kudos [?]: 279 [0], given: 7

Schools: University of Chicago, Wharton School
Re: DS: Is A positive? [#permalink]

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19 Sep 2007, 09:50
b14kumar wrote:
b14kumar wrote:
Fistail wrote:
GK_Gmat wrote:
Is A positive?

1) x^2 - 2x +A is positive for all x
2) A*x^2 + 1 is positive for all x

Pls. explain. Thanks.

1: x^2 - 2x +A > 0

refrese the inequality as : x^2 - 2x + 1 + A -1 > 0
so it is reduced to (x - 1)^2 + A – 1 > 0.
now (x -1)^2 can be 0 or grater than 0. if it is 0, A - 1 has to be +ve and to be so, A has to be grater than 1. so suff.

2: A*x^2 + 1 is positive for all x.

it is clearly insufficient because x^2 is 0, A could be +ve or -ve and the expression still is +ve..

so A works here.

Indeed a good approach Fistail.

But how about putting the value of x = -1 and obtaining the possible values of A as I mentioned in my above earlier post?

Writing again:

ST1: x^2 - 2x +A is positive for all x

Let's say x = -1

Then, 1 + 2 + A > 0
=> 3 + A > 0
=> A > (-3)

So A may be -2 , -1 , 0 or > 0

Hence A may be positive or negative.

- Brajesh

I again looked at your approach.

As per you:

refrese the inequality as : x^2 - 2x + 1 + A -1 > 0
so it is reduced to (x - 1)^2 + A – 1 > 0.
now (x -1)^2 can be 0 or grater than 0.
if it is 0, A - 1 has to be +ve and to be so, A has to be grater than 1. so suff.
Why are you assuming only the case when (x-1)^2 is equal to 0?

Well, (x -1)^2 will always be >= 0 but it does not mean that "A – 1" has to be positive for all the cases.
Imagine, (x -1)^2 is equal to 4 (by taking x = 3 ) , in this case, (A-1) can be (-3) i.e A can be (-2) and still the whole inequality will be intact.

- Brajesh

go for the extream case. make (x^2 - 1) minimum. for other values of (x - 1)^2, its not a problem but when (x - 1)^2 is lowest what happens to the value of A is important? since A must be +ve when (x - 1)^2 is minimum, the A is +ve.

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20 Sep 2007, 01:39
At first I thought E, but then took a step back and realized St1 is sufficient. St2 is not sufficient in itself since A could equal say -0.5 and the expression would still be positive for all x.

Kudos [?]: 9 [0], given: 0

20 Sep 2007, 01:39

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