Simplify k^2-3>=0

$k^{2} - 3 \geq 0$

Step 1

Add $3$ to both sides of the inequality.

$k^{2} \geq 3$

Step 2

Take the specified root of both sides of the inequality to eliminate the exponent on the left side.

$\sqrt{k^{2}} \geq \sqrt{3}$

Step 3

Simplify the left side.

Pull terms out from under the radical.

$| k | \geq \sqrt{3}$

Step 4

Write $| k | \geq \sqrt{3}$ as a piecewise.

To find the interval for the first piece, find where the inside of the absolute value is non-negative.

$k \geq 0$

In the piece where $k$ is non-negative, remove the absolute value.

$k \geq \sqrt{3}$

To find the interval for the second piece, find where the inside of the absolute value is negative.

$k < 0$

In the piece where $k$ is negative, remove the absolute value and multiply by $- 1$ .

$- k \geq \sqrt{3}$

Write as a piecewise.

${\begin{cases} k \geq \sqrt{3} & k \geq 0 \\ - k \geq \sqrt{3} & k < 0 \end{cases}$

Step 5

Find the intersection of $k \geq \sqrt{3}$ and $k \geq 0$ .

$k \geq \sqrt{3}$

Step 6

Divide each term in $- k \geq \sqrt{3}$ by $- 1$ and simplify.

Divide each term in $- k \geq \sqrt{3}$ by $- 1$ . When multiplying or dividing both sides of an inequality by a negative value, flip the direction of the inequality sign.

$\frac{- k}{- 1} \leq \frac{\sqrt{3}}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{k}{1} \leq \frac{\sqrt{3}}{- 1}$

Divide $k$ by $1$ .

$k \leq \frac{\sqrt{3}}{- 1}$

Simplify the right side.

Move the negative one from the denominator of $\frac{\sqrt{3}}{- 1}$ .

$k \leq - 1 \cdot \sqrt{3}$

Rewrite $- 1 \cdot \sqrt{3}$ as $- \sqrt{3}$ .

$k \leq - \sqrt{3}$

Step 7

Find the union of the solutions.

$k \leq - \sqrt{3}$ or $k \geq \sqrt{3}$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$k \leq - \sqrt{3} or k \geq \sqrt{3}$

Interval Notation:

$(- \infty, - \sqrt{3}] \cup [\sqrt{3}, \infty)$

Step 9

Simplify k^2-3>=0

Related Questions