Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized for You

we will pick new questions that match your level based on your Timer History

Track Your Progress

every week, we’ll send you an estimated GMAT score based on your performance

Practice Pays

we will pick new questions that match your level based on your Timer History

Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.

It appears that you are browsing the GMAT Club forum unregistered!

Signing up is free, quick, and confidential.
Join other 500,000 members and get the full benefits of GMAT Club

Registration gives you:

Tests

Take 11 tests and quizzes from GMAT Club and leading GMAT prep companies such as Manhattan GMAT,
Knewton, and others. All are free for GMAT Club members.

Applicant Stats

View detailed applicant stats such as GPA, GMAT score, work experience, location, application
status, and more

Books/Downloads

Download thousands of study notes,
question collections, GMAT Club’s
Grammar and Math books.
All are free!

Thank you for using the timer!
We noticed you are actually not timing your practice. Click the START button first next time you use the timer.
There are many benefits to timing your practice, including:

x^2-2x+A is positive for all x Ax^2+1 is positive for all x

OA is A

Is \(A>0\)?

(1) \(x^2-2x+A\) is positive for all \(x\):

Quadratic expression \(x^2-2x+A\) is a function of of upward parabola (it's upward as coefficient of \(x^2\) is positive). We are told that this expression is positive for all \(x\) --> \(x^2-2x+A>0\), which means that this parabola is "above" X-axis OR in other words parabola has no intersections with X-axis OR equation \(x^2-2x+A=0\) has no real roots.

Quadratic equation to has no real roots discriminant must be negative --> \(D=2^2-4A=4-4A<0\) --> \(1-A<0\) --> \(A>1\).

Sufficient.

(2) \(Ax^2+1\) is positive for all \(x\):

\(Ax^2+1>0\) --> when \(A\geq0\) this expression is positive for all \(x\). So \(A\) can be zero too.

I dont get it sorry... I mean I understand your equations Bunuel, but I tried first with picking numbers:

If I pick -0.5 for x --> x^2-2x+A>0 will hold for A > -1.25

...

Where is my mistake??

The point here is that \(x^2-2x+A>0\) for all \(x-es\).

Let's do this in another way:

We have \((x^2-2x)+A>0\) for all \(x-es\). The sum of 2 quantities (\(x^2-2x\) and \(A\)) is positive for all \(x-es\). So for the least value of \(x^2-2x\), \(A\) must make the whole expression positive.

So what is the least value of \(x^2-2x\)? The least value of quadratic expression \(ax^2+bx+c\) is when \(x=-\frac{b}{2a}\), so in our case the least value of \(x^2-2x\) is when \(x=-\frac{-2}{2}=1\) --> \(x^2-2x=-1\) --> \(-1+A>0\) --> \(A>1\).

OR:

\(x^2-2x+A>0\) --> \(x^2-2x+1+A-1>0\) --> \((x-1)^2+A-1>0\) --> least value of \((x-1)^2\) is zero thus \(A-1\) must be positive (\(0+A-1>0\))--> \(A-1>0\) --> \(A>1\).

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

\(x^2 - 2x + A = 9 +6 +A = 15 + A >0\)

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all \(x-es\) means that no matter what \(x\) you pick \(x^2 - 2x + A\) must be positive. So it must be positive even for the lowest value of \(x^2 - 2x\) which is -1 --> so \(-1+A\) must be positive hence A must be more than 1.

Now again: if A>1 then for any \(x\) expression \(x^2 - 2x + A\) is positive.

But if A=-15 (or any other number less than 1) we can find some \(x-es\) for which expression \(x^2 - 2x + A\) is not positive, so theese values of A (values of \(A\leq{1}\)) are not valid.

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

\(x^2 - 2x + A = 9 +6 +A = 15 + A >0\)

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all \(x-es\) means that no matter what \(x\) you pick \(x^2 - 2x + A\) must be positive. So it must be positive even for the lowest value of \(x^2 - 2x\) which is -1 --> so \(-1+A\) must be positive hence A must be more than 1.

Now again: if A>1 then for any \(x\) expression \(x^2 - 2x + A\) is positive.

But if A=-15 (or any other number less than 1) we can find some \(x-es\) for which expression \(x^2 - 2x + A\) is not positive, so theese values of A (values of \(A\leq{1}\)) are not valid.

Hope it's clear.

Thanks for the reply Bunuel,

I understood the approach but the fact which is baffling me is that say the equation after subsituting value of x=-3 i.e 15+ A > 0 now we can have a value of A=-3 or may be -4 etc and still have the value of the equation in statement 1 as +ve

Am I thinking too much or just lacking some thing basic concept.

I really liked approached here but I still have some confusion,

Say for e.g if try to pick the numbers say x = -3

Then the equation in the first statement becomes

\(x^2 - 2x + A = 9 +6 +A = 15 + A >0\)

So now if we see A can have -ve and +ve values, isnt it ?

I am confused with this.

Please explain, whats wrong with this one.

Cheers

Not every question can be solved by number picking.

For all \(x-es\) means that no matter what \(x\) you pick \(x^2 - 2x + A\) must be positive. So it must be positive even for the lowest value of \(x^2 - 2x\) which is -1 --> so \(-1+A\) must be positive hence A must be more than 1.

Now again: if A>1 then for any \(x\) expression \(x^2 - 2x + A\) is positive.

But if A=-15 (or any other number less than 1) we can find some \(x-es\) for which expression \(x^2 - 2x + A\) is not positive, so theese values of A (values of \(A\leq{1}\)) are not valid.

Hope it's clear.

Thanks for the reply Bunuel,

I understood the approach but the fact which is baffling me is that say the equation after subsituting value of x=-3 i.e 15+ A > 0 now we can have a value of A=-3 or may be -4 etc and still have the value of the equation in statement 1 as +ve

Am I thinking too much or just lacking some thing basic concept.

I appreciate your patience.

I think you just don't understand one thing in statement (1): \(x^2-2x+A>0\) FOR ALL \(x-es\).

You say that if \(x=-3\) then \(A\) can be for example -10 (or any number more than -15) and \(x^2-2x+A\) will be positive, \(but\) if \(x=1\) does \(A=-10\) makes \(x^2-2x+A\) positive? NO!

So you should find such value of \(A\) (such range) for which \(x^2-2x+A\) is positive no matter what value of \(x\) you'll plug. And the way how to find this range is shown in my previous posts. _________________

1) X^2-2X+A is positive for all X I think A could not be the answer, for example, if A = 0, and X = 4, then also the expression is positive, but A = 0 is neither positive nor negative Again, if A = 1, and and X = 4, then also the expression is positive 2) AX^2 + 1 is positive for all X Same logic as above, if A is 0, then the expression is positive, and the expression is also postive for any value of X where A > 0

In a nutshell, I too think the answer is E. _________________

Formula of Life -> Achievement/Potential = k * Happiness (where k is a constant)

1) X^2-2X+A is positive for all X 2) AX^2 + 1 is positive for all X

given answer as A...but i thought it should be E.. source: hard problems from gmatclub tests number properties I

1) X^2-2X+A is positive for all X

For all values of X,\(X^2-2X+A > 0\) This means, for X = 0, \(X^2-2X+A > 0\); for X = 1, \(X^2-2X+A > 0\); for X = -2, \(X^2-2X+A > 0\) etc etc etc

Let's put X = 0. \(0^2-2*0+A > 0\) should hold. Therefore, A > 0 should hold. Sufficient.

2) AX^2 + 1 is positive for all X

For all X, \(AX^2 + 1 > 0\) Here, A could be positive or A could be 0 (since, when A = 0, we get 1 > 0 which holds no matter what the value of X.) Since A can be 0, we cannot say whether A is positive. Not Sufficient.

x^2-2x+A is positive for all x Ax^2+1 is positive for all x

OA is A

(2) \(Ax^2+1\) is positive for all \(x\):

\(Ax^2+1>0\) --> when \(A\geq0\) this expression is positive for all \(x\). So \(A\) can be zero too.

Not sufficient.

Answer: A.

Why didn't you use the discriminant formula to assess statement 2?

I tried the discriminant rule and got a>0. I had 0-4a<0 which turns to a>0.

What am I missing here?

Thanks, Diana

You are right: if we use the same approach for (2) then we'll get A>0 BUT if A=0 then Ax^2+1 won't be a quadratic function anymore. So this approach will work only if A doesn't equal to zero, but we can not eliminate this case and if A=0 then Ax^2+1=1 is also always positive. Hence Ax^2+1 is positive for A>0 (if we use quadratic function approach) as well as for A=0, so for \(A\geq0\).

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

I got E and my way of solving is as below: St1. x^2 - 2x +A > 0 Let x=0, so A>0. Let x=-1, so A>-3. In this case A can be negative or positive. Insufficient.

St2. A*x^2 +1 > 0 Let x=-1, so A>-1. Again A can be positive or negative. Insufficient.

St1+St2: Let x=-1, so A > -1. Again A can be positive or negative. Insufficient.

So it's E. However the OA is not E. Please advise.

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

I got E and my way of solving is as below: St1. x^2 - 2x +A > 0 Let x=0, so A>0. Let x=-1, so A>-3. In this case A can be negative or positive. Insufficient.

St2. A*x^2 +1 > 0 Let x=-1, so A>-1. Again A can be positive or negative. Insufficient.

St1+St2: Let x=-1, so A > -1. Again A can be positive or negative. Insufficient.

So it's E. However the OA is not E. Please advise.

Merging similar topics. Please refer to the solutions above. _________________

Re: Is A positive? x^2-2x+A is positive for all x Ax^2+1 is [#permalink]

Show Tags

21 Aug 2012, 16:51

1

This post received KUDOS

best way to deal this problem is to bet on A more than X.. it wud b yes if A>0 Or No ,if A<0 .... then first assume A>0 , then check whether statement 1 & 2 is true or not for all value of X.... then assume A<0 ,then check whether statement 1 & 2 is true or not for all value of X.... _________________

So, my final tally is in. I applied to three b schools in total this season: INSEAD – admitted MIT Sloan – admitted Wharton – waitlisted and dinged No...

HBS alum talks about effective altruism and founding and ultimately closing MBAs Across America at TED: Casey Gerald speaks at TED2016 – Dream, February 15-19, 2016, Vancouver Convention Center...

By Libby Koerbel Engaging a room of more than 100 people for two straight hours is no easy task, but the Women’s Business Association (WBA), Professor Victoria Medvec...