If |x| < x^2, which of the following must be true ?

NandishSS wrote:If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

|x| < x² implies that x is NONZERO.

Since x is nonzero, |x| > 0 and x² > 0.
Implication:
Both sides of the inequality are positive, allowing us to SQUARE the inequality.
(|x|)² < (x²)²
x² < x�.

Since x² > 0, we can safely DIVIDE both sides by x²:
x²/x² < x�/x²
1 < x²
x² > 1.

The correct answer is E.

Another approach:

For any value of z, |z| = √z².

We're told that |x| < x², so √x² < x².

We know that x² > 0 (if it were equal to 0, then x² wouldn't be > √x²). That means we can safely divide both sides of the inequality by √x² ...

√x² < x², then divide both sides:

1 < √x²

We know that √x² < x², so making the two inequalities into a chain gives

1 < √x² < x², or 1 < x², so E.

NandishSS wrote:If |x| < x^2, which of the following must be true ?

A. x > 0
B. x < 0
C. x > 1
D. -1 < x < 1
E. x^2 > 1

If x = 0, then |0| will not be less than 0^2 since they both are equal to 0. Thus, we know x can’t be 0.

Since |x| = x when x is positive and -x when x is negative, let’s rewrite the inequality without the absolute value sign. That is, if x > 0, then we have x < x^2, and if x < 0, then we have -x < x^2.

Case 1: If x > 0,

x < x^2

Dividing both sides by x, we have:

1 < x

Case 2: If x < 0,

-x < x^2

Dividing both sides by x (switching the inequality sign since x is negative), we have:

-1 > x

Thus, we have x > 1 or x < -1; in that case, x^2 must be greater than 1.

Answer: E