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Sorry, hopefully it's a badly worded question - but if we're asking if statement one definitively says whether A is positive or not, I'm not sure about the answer!

statement 1) says x^2 - 2x + A must be positive... inserting x=3.... (3^2) - (3*2) + A must be positive... 9 - 6 + A must be positive 3 + A must be positive

A can be -2, -1, 0 and any value greater than 0... in usual DS-type questions, this suggests 'insufficient' to me...

as for statement 2) A(x^2) + 1 must be positive.... we know x^2 will be positive for any value of x, but since A can be 0 and still satisfy this statement, this is insufficient also.

If someone can explain why the range of values A can take where x=3 in statement 1) definitively answers the 'is A positive?' question, it'd be really appreciated _________________

Sorry, hopefully it's a badly worded question - but if we're asking if statement one definitively says whether A is positive or not, I'm not sure about the answer!

statement 1) says x^2 - 2x + A must be positive... inserting x=3.... (3^2) - (3*2) + A must be positive... 9 - 6 + A must be positive 3 + A must be positive

A can be -2, -1, 0 and any value greater than 0... in usual DS-type questions, this suggests 'insufficient' to me...

as for statement 2) A(x^2) + 1 must be positive.... we know x^2 will be positive for any value of x, but since A can be 0 and still satisfy this statement, this is insufficient also.

If someone can explain why the range of values A can take where x=3 in statement 1) definitively answers the 'is A positive?' question, it'd be really appreciated

Question is correct IMO. x is the variable and A a constant. St1 says that A is some magical value such that the expression will always be +ve for any x. No matter what value you assign to x, the expression will always be +ve.

Say, A=-2 For x=3; the expression is +ve. Great!!! for x=1; the expression is -ve(falsifies the statement) Means; A must not be -2. _________________

St1 says that A is some magical value such that the expression will always be +ve for any x. No matter what value you assign to x, the expression will always be +ve.

thanks - error in interpretation (treating 'a' as variable that will alter to satisfy the equation accordingly and not a constant). _________________

As I understood...minimum value of A which makes the expression xsq2-2*x+A for all positive. If it's given that x2-x*x+a> 0 for all x , then we must be sure that is such that the inequality holds true.

If we know that a>1 , we can answer the question ( is positive).

After reading this, I definitely see why stmt 1 is sufficient and stmt 2 is not. However, I am wondering why can't we evaluate statement 2 by finding the range of the discriminant, as we did with statement 1. I tried doing so and came up with A>0, but we know that A is greater than or equal to 0.

Bunuel wrote:

CrushTheGMAT wrote:

Is A positive?

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

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0. The correct answer is A.

***** I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is A>0?

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

f(x)=x^2-2x+A is a function of of upward parabola (as coefficient of x^2 is positive). We are told that it's positive for all x --> f(x)=x^2-2x+A>0, which means that this function 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.

In statement 1 why can't we consider x = 10000 (some insane large number), if we do then we can the square of x will be much greater than -2x and in that case I think we can have A = -1 and still have a positive value for the expression

I still believe statement 1 is insuff. I don't get it why suff?

In statement 1 why can't we consider x = 10000 (some insane large number), if we do then we can the square of x will be much greater than -2x and in that case I think we can have A = -1 and still have a positive value for the expression

I still believe statement 1 is insuff. I don't get it why suff?

Sure you can find SOME x for which A will be negative, BUT (1) says that x^2-2x+A is positive FOR ALL x-es. This expression to be positive for ALL x-es A must be more than 1. Refer to my post here to see why: m13-q23-70269.html#p716027

f(x)=x^2-2x+A is a function of of upward parabola (as coefficient of x^2 is positive). We are told that it's positive for all x --> f(x)=x^2-2x+A>0, which means that this function 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.

Not sufficient.

Answer: A.

After reading this, I definitely see why stmt 1 is sufficient and stmt 2 is not. However, I am wondering why can't we evaluate statement 2 by finding the range of the discriminant, as we did with statement 1. I tried doing so and came up with A>0, but we know that A is greater than or equal to 0.

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.

why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

Teal, please read the threads you are replying to carefully.

S1 is sufficient. x^2 - 2*x + A = (x - 1)^2 + (A - 1). For this expression to be always positive A - 1 has to be more than 0. A has to be more than 1. S2 is not sufficient. A*x^2 + 1 is positive for all x even if A is 0. The correct answer is A.

***** I believe the solution is incorrect on this question:

S1 does not work. For example, use x=10. The equation yields 10^2-2*10+A>0, so 80+A>0. Than means that A>-80, and hence can be negative or positive. S1 says A>-x^2+2x, and if you graph the function the answer becomes obvious.

S2 does not work, and neither does S1+S2, so the answer should be E.

Question: is A>0?

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

f(x)=x^2-2x+A is a function of of upward parabola (as coefficient of x^2 is positive). We are told that it's positive for all x --> f(x)=x^2-2x+A>0, which means that this function 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.

why can't we apply the negative discriminant logic in statement 2; the parabola in stat 2 will also not cut x axis ( in all possible values of a, which only defines the slope of parabola). But discriminant would be

0-4*a*1<0

But that does not give us 0 for A and lead us to suff. Why does this not work in stat 2?

hi,

a general quadratic equation is like this:

y=ax^2+bx+c

now your doubt is "why can't we apply the negative discriminant logic in statement 2".

you can apply the logic of negative discriminant only when you are sure that coefficient of x^2 is positive...i.e "a" is positive.

now in our question equation is AX^2+1

here we dont know whether A is positive or negative now if we use the discriminant rule in this then we are already assuming that "A" is positive...

and we are neglecting the possibility of A < 0...

so TAKEAWAYS:

use the discriminant concept only when coefficient of X^2 is positive.

hope it helps.

SKM _________________

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