Mo2men wrote:Is A positive?

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

Statement 1:
In other words, the graph of y = x² - 2x + A lies entirely above the x-axis, so that the value of y is always positive.

A parabola of the form y = ax² + bx + c, where a>0, opens UPWARD.
The result is a U-shaped graph that looks like this:
U.
The DISCRIMINANT of the parabola is equal to b² - 4ac.
The U-shaped graph will lie entirely above the x-axis -- and thus will yield only positive values for y -- if its discriminant is negative.
Implication:
Since y = x² - 2x + A must yield only positive values for y, its discriminant must be negative.

In y = x² - 2x + A, a=1, b=-2, and c=A.
Since b² - 4ac < 0, we get:
(-2)² - 4*1*A < 0
4 - 4A < 0
-4A < -4
A > 1.
SUFFICIENT.

Statement 2:
In other words, the graph of y = Ax² + 1 lies entirely above the x-axis, so that the value of y is always positive.

Case 1: A=0, so that y = Ax² + 1 becomes y=1.
Here, the graph is a horizontal line that lie entirely above the x-axis.

Case 2: A=1, so that y = Ax² + 1 becomes y = x² + 1
Here, since x² cannot be negative, every value for y will be positive, yielding a graph that lies entirely above the x-axis.

Since A=0 in Case 1, but A>0 in Case 2, INSUFFICIENT.

The correct answer is A.

GMATGuruNY wrote:

Mo2men wrote:Is A positive?

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

Statement 1:
In other words, the graph of y = x² - 2x + A lies entirely above the x-axis, so that the value of y is always positive.

A parabola of the form y = ax² + bx + c, where a>0, opens UPWARD.
The result is a U-shaped graph that looks like this:
U.
The DISCRIMINANT of the parabola is equal to b² - 4ac.
The U-shaped graph will lie entirely above the x-axis -- and thus will yield only positive values for y -- if its discriminant is negative.
Implication:
Since y = x² - 2x + A must yield only positive values for y, its discriminant must be negative.

In y = x² - 2x + A, a=1, b=-2, and c=A.
Since b² - 4ac < 0, we get:
(-2)² - 4*1*A < 0
4 - 4A < 0
-4A < -4
A > 1.
SUFFICIENT.

Dear Mitch,
If I know the basics which you mentioned about parabola. Do I need to proceed to calculate the value of the DISCRIMINANT??

Thanks

Mo2men wrote: Dear Mitch,
If I know the basics which you mentioned about parabola. Do I need to proceed to calculate the value of the DISCRIMINANT??

Thanks

Since the value of A is unknown -- Statement 1 implies only that A>0 -- the value of the discriminant cannot be determined.

GMATGuruNY wrote:

Mo2men wrote: Dear Mitch,
If I know the basics which you mentioned about parabola. Do I need to proceed to calculate the value of the DISCRIMINANT??

Thanks

Since the value of A is unknown -- Statement 1 implies only that A>0 -- the value of the discriminant cannot be determined.

Dear Mitch,

I may not have phrased my question properly.

In you solution for statement 1:

In other words, the graph of y = x² - 2x + A lies entirely above the x-axis, so that the value of y is always positive.

A parabola of the form y = ax² + bx + c, where a>0, opens UPWARD.
The result is a U-shaped graph that looks like this:
U.

Is above information are enough to conclude that A is positive?

I want to avoid the DISCRIMINANT and formulas.

Thanks

Mo2men wrote:

GMATGuruNY wrote:

Mo2men wrote: Dear Mitch,
If I know the basics which you mentioned about parabola. Do I need to proceed to calculate the value of the DISCRIMINANT??

Thanks

Since the value of A is unknown -- Statement 1 implies only that A>0 -- the value of the discriminant cannot be determined.

Dear Mitch,

I may not have phrased my question properly.

In you solution for statement 1:

In other words, the graph of y = x² - 2x + A lies entirely above the x-axis, so that the value of y is always positive.

A parabola of the form y = ax² + bx + c, where a>0, opens UPWARD.
The result is a U-shaped graph that looks like this:
U.

Is above information are enough to conclude that A is positive?

I want to avoid the DISCRIMINANT and formulas.

Thanks

Yes.
To guarantee that the U-shaped graph yielded by y = x² ± bx + c lies above the x-axis, the value of c must be positive.
Thus:
To guarantee that the U-shaped graph yielded by y = x² - 2x + A lies above the x-axis, the value of A must be positive.