Square Root Inequality

Algebra Level 3

Which of the following is equivalent to

$-1 < \sqrt{x} < 2 ?$

-1 < x < 4

0 < x < 4

1 < x < 4

0 \leq x < 4

1 solution

Micah Wood
Nov 3, 2016

If $x$ is under $0$ , we would be comparing imaginary number to $-1$ , which is nonsense, so we can restrict $\sqrt x$ to $0\le \sqrt x < 2$ . We can square all sides, giving us $\boxed{0\le x < 4}$

We cannot just blindly square everything in an inequality.

Squaring of positive numbers work because we are using $a > b >0 \Rightarrow a^2 > ab > b^2$ .

If we square a negative number, we are multiplying by a negative number and hence need to change the sign. So if $a > 0 > b$ , then all that we have is $a^2 > ab < b^2$ so we cannot compare $a^2$ with $b^2$ .

Chung Kevin - 4 years, 7 months ago

Of course. For domain of $\mathbb R$ , $\sqrt x$ is only defined for $x\ge 0$ and hence range must be great than or equal to 0. So $\sqrt x$ cannot be negative number.

There is no negative number to deal with, so we can square everything in inequality.

Micah Wood - 4 years, 7 months ago

I was thinking along the lines of $- 1 < x < 2 \Rightarrow 0 \leq x^2 < 4$ , so this avoids just using the domain of $\sqrt{x}$ as the restriction.

Do you think the problem would be improved in this way?

Chung Kevin - 4 years, 7 months ago

Square Root Inequality

1 solution

0 pending reports