infiniti007 wrote:Is |xy| > x²y²?

1.) 0 < x² < 1/4
2.) 0 < y² < 1/9

|xy| > x²y² only if x and y are NONZERO, with the result that both sides of the inequality are POSITIVE.
Since |xy| > x²y² only if both sides are positive, we can safely square the inequality:
(|xy|)² > (x²y²)²
x²y² > x�y�.

Since x and y are nonzero, x²y²>0.
Thus, we can safely divide each side by x²y²:
x²y²/x²y² > x�y�/x²y²
1 > x²y².

Question stem, rephrased:
Is x²y² < 1?

Statement 1: 0 < x² < 1/4
If x² = 1/10 and y² = 1/10, then x²y² < 1.
If x² = 1/10 and y² = 100, then x²y² > 1.
INSUFFICIENT.

Statement 2: 0 < y² < 1/9
If y² = 1/10 and x² = 1/10, then x²y² < 1.
If y² = 1/10 and x² = 100, then x²y² > 1.
INSUFFICIENT.

Statements combined:
Since x² and y² are both POSITIVE FRACTIONS, x²y² < 1.
SUFFICIENT.

The correct answer is C.

|xy| > x²y² only if x and y are NONZERO, with the result that both sides of the inequality are POSITIVE.
Since |xy| > x²y² only if both sides are positive, we can safely square the inequality:
(|xy|)² > (x²y²)²
x²y² > x�y�.

I may be understanding this incorrectly. I understand the inequality would not hold if x and y are each equal to 0. What if x and y were each equal to -1? Or what about x and y each equal to 1? Wouldn't these values not hold also?

infiniti007 wrote:

|xy| > x²y² only if x and y are NONZERO, with the result that both sides of the inequality are POSITIVE.
Since |xy| > x²y² only if both sides are positive, we can safely square the inequality:
(|xy|)² > (x²y²)²
x²y² > x�y�.

I may be understanding this incorrectly. I understand the inequality would not hold if x and y are each equal to 0. What if x and y were each equal to -1? Or what about x and y each equal to 1? Wouldn't these values not hold also?

Our concern is not the value of each side but the SIGN of each side.

Example: x>y
Here, the signs of x and y are unknown.
For this reason, we CANNOT simply square both sides and conclude that x² > y².
If x=1 and y=-10, then x>y but x² < y².

Problem above: |xy| > x²y²
Here, |xy| > x²y² requires that both x and y be NONZERO.
As a result, |xy| > 0 and x²y² > 0.
Since both sides of the inequality must be POSITIVE, we can safely square the inequality.

Hello,

I have a doubt:

is it true |x| = root (x)


Thanks
Vishal

vishalwin wrote:Hello,

I have a doubt:

is it true |x| = root (x)


Thanks
Vishal

|x| = √(x²).
The reason is that √ means the POSITIVE ROOT ONLY.

Example: x=-4
|x| = |-4| = 4.
√(x²) = √((-4)²) = √16 = the positive square root of 16 = 4.
Thus:
|x| = √(x²).