kaps786 wrote:Is x>0

1. x<x^2

2. x<x^3

Can someone show me how to solve quadratic inequalities particularly how to identify the solutions on the number line.

OA IS E

We can use the picking numbers approach here.

(1) x < x²
If x = -2, x² = (-2)² = 4, then x < 0.
If x = 2, x² = (2)² = 4, then x > 0.
If x = -1/2, x² = (-1/2)² = 1/4, then x < 0.
No definite answer from the above examples. So statement 1 is NOT sufficient.

(2) x < x^3
If x = 2, x^3 = (2)^3 = 8, then x > 0.
If x = -1/2, x^3 = (-1/2)^3 = -1/8, then x < 0.
No definite answer from the above examples. So statement 2 is NOT sufficient.

Combining (1) and (2), if we take the same examples as taken in statements 1 and 2, x = 2, x = -1/2, then again we do not get a definite answer.

The correct answer is E.

goelmaya wrote:Oh! i got my mistake..

x(x-1) > 0
=> x > 0 and x-1>0 => x > 1

or x<0 and x-1 <0 => x < 0

We can always check the solution:
x < 0: Let us take x = -1, then x² - x = 1 + 1 = 2 > 0
x > 1: Let us take x = 2, then x² - x = 4 - 2 = 2 > 0

goelmaya wrote:Thanks a lot

You are welcome