An algebra problem by Ivander Jonathan

Algebra Level 2

Given that $x^2 \le x$ . How many values of x satisfies the equation?

1 0

\infty

4 solutions

Suklo Charan Kisku
Feb 13, 2015

The solution set is [0,1]

Albert Go
Feb 13, 2015

divide both sides by x. Then you'll get x<1, since there are infinite numbers below 1, the answer therefore is infinite

Vishal S
Feb 5, 2015

$x^2$ $\leq x$ $\Rightarrow$ $\frac {x^2}{x}$ $\leq 1$ $\Rightarrow$ $x^2 \leq 1$ .

Since there are infinite numbers less than 1.Therefore the answer is $\infty$

But the square of negative numbers is positive. So how are they less than 1. Please answer!!!!!!!!

Ankit Kumar Jain - 6 years, 4 months ago

I did not tell the value of x is negative.It may be any real number.

For example (1/2)^2 < 1/2.

Vishal S - 6 years, 4 months ago

Ok..... Thanks

Ankit Kumar Jain - 6 years, 4 months ago

Shouldn't the last step be x<1 ?

Since you divided x^2 by x

Vasudev Chandna - 6 years, 2 months ago

Niranjan Khanderia
Feb 4, 2015

All + tive real number less or equal to 1 and 0 satisfies the equation. Hence they are infinity.