Calculus Examples

$x^{2} + 3 x - 1 \geq 0$

Step 1

Convert the inequality to an equation.

$x^{2} + 3 x - 1 = 0$

Step 2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 3

Substitute the values $a = 1$ , $b = 3$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

$\frac{- 3 \pm \sqrt{3^{2} - 4 \cdot (1 \cdot - 1)}}{2 \cdot 1}$

Step 4

Simplify.

Tap for more steps...

Step 4.1

Simplify the numerator.

Tap for more steps...

Step 4.1.1

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 4.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

Tap for more steps...

Step 4.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 1}}{2 \cdot 1}$

Step 4.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2 \cdot 1}$

Step 4.1.3

Add $9$ and $4$ .

$x = \frac{- 3 \pm \sqrt{13}}{2 \cdot 1}$

Step 4.2

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{13}}{2}$

Step 5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

Tap for more steps...

Step 5.1

Simplify the numerator.

Tap for more steps...

Step 5.1.1

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 5.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

Tap for more steps...

Step 5.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 1}}{2 \cdot 1}$

Step 5.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2 \cdot 1}$

Step 5.1.3

Add $9$ and $4$ .

$x = \frac{- 3 \pm \sqrt{13}}{2 \cdot 1}$

Step 5.2

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{13}}{2}$

Step 5.3

Change the $\pm$ to $+$ .

$x = \frac{- 3 + \sqrt{13}}{2}$

Step 5.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 + \sqrt{13}}{2}$

Step 5.5

Factor $- 1$ out of $\sqrt{13}$ .

$x = \frac{- 1 \cdot 3 - 1 (- \sqrt{13})}{2}$

Step 5.6

Factor $- 1$ out of $- 1 (3) - 1 (- \sqrt{13})$ .

$x = \frac{- 1 (3 - \sqrt{13})}{2}$

Step 5.7

Move the negative in front of the fraction.

$x = - \frac{3 - \sqrt{13}}{2}$

Step 6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

Tap for more steps...

Step 6.1

Simplify the numerator.

Tap for more steps...

Step 6.1.1

Raise $3$ to the power of $2$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot 1 \cdot - 1}}{2 \cdot 1}$

Step 6.1.2

Multiply $- 4 \cdot 1 \cdot - 1$ .

Tap for more steps...

Step 6.1.2.1

Multiply $- 4$ by $1$ .

$x = \frac{- 3 \pm \sqrt{9 - 4 \cdot - 1}}{2 \cdot 1}$

Step 6.1.2.2

Multiply $- 4$ by $- 1$ .

$x = \frac{- 3 \pm \sqrt{9 + 4}}{2 \cdot 1}$

Step 6.1.3

Add $9$ and $4$ .

$x = \frac{- 3 \pm \sqrt{13}}{2 \cdot 1}$

Step 6.2

Multiply $2$ by $1$ .

$x = \frac{- 3 \pm \sqrt{13}}{2}$

Step 6.3

Change the $\pm$ to $-$ .

$x = \frac{- 3 - \sqrt{13}}{2}$

Step 6.4

Rewrite $- 3$ as $- 1 (3)$ .

$x = \frac{- 1 \cdot 3 - \sqrt{13}}{2}$

Step 6.5

Factor $- 1$ out of $- \sqrt{13}$ .

$x = \frac{- 1 \cdot 3 - (\sqrt{13})}{2}$

Step 6.6

Factor $- 1$ out of $- 1 (3) - (\sqrt{13})$ .

$x = \frac{- 1 (3 + \sqrt{13})}{2}$

Step 6.7

Move the negative in front of the fraction.

$x = - \frac{3 + \sqrt{13}}{2}$

Step 7

Consolidate the solutions.

$x = - \frac{3 - \sqrt{13}}{2}, - \frac{3 + \sqrt{13}}{2}$

Step 8

Use each root to create test intervals.

$x < - \frac{3 + \sqrt{13}}{2}$

$- \frac{3 + \sqrt{13}}{2} < x < - \frac{3 - \sqrt{13}}{2}$

$x > - \frac{3 - \sqrt{13}}{2}$

Step 9

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

Tap for more steps...

Step 9.1

Test a value on the interval $x < - \frac{3 + \sqrt{13}}{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 9.1.1

Choose a value on the interval $x < - \frac{3 + \sqrt{13}}{2}$ and see if this value makes the original inequality true.

$x = - 6$

Step 9.1.2

Replace $x$ with $- 6$ in the original inequality.

${(- 6)}^{2} + 3 (- 6) - 1 \geq 0$

Step 9.1.3

The left side $17$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.2

Test a value on the interval $- \frac{3 + \sqrt{13}}{2} < x < - \frac{3 - \sqrt{13}}{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 9.2.1

Choose a value on the interval $- \frac{3 + \sqrt{13}}{2} < x < - \frac{3 - \sqrt{13}}{2}$ and see if this value makes the original inequality true.

$x = 0$

Step 9.2.2

Replace $x$ with $0$ in the original inequality.

${(0)}^{2} + 3 (0) - 1 \geq 0$

Step 9.2.3

The left side $- 1$ is less than the right side $0$ , which means that the given statement is false.

False

Step 9.3

Test a value on the interval $x > - \frac{3 - \sqrt{13}}{2}$ to see if it makes the inequality true.

Tap for more steps...

Step 9.3.1

Choose a value on the interval $x > - \frac{3 - \sqrt{13}}{2}$ and see if this value makes the original inequality true.

$x = 3$

Step 9.3.2

Replace $x$ with $3$ in the original inequality.

${(3)}^{2} + 3 (3) - 1 \geq 0$

Step 9.3.3

The left side $17$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 9.4

Compare the intervals to determine which ones satisfy the original inequality.

$x < - \frac{3 + \sqrt{13}}{2}$ True

$- \frac{3 + \sqrt{13}}{2} < x < - \frac{3 - \sqrt{13}}{2}$ False

$x > - \frac{3 - \sqrt{13}}{2}$ True

$x < - \frac{3 + \sqrt{13}}{2}$ True

$- \frac{3 + \sqrt{13}}{2} < x < - \frac{3 - \sqrt{13}}{2}$ False

$x > - \frac{3 - \sqrt{13}}{2}$ True

Step 10

The solution consists of all of the true intervals.

$x \leq - \frac{3 + \sqrt{13}}{2}$ or $x \geq - \frac{3 - \sqrt{13}}{2}$

Step 11

Convert the inequality to interval notation.

$(- \infty, - \frac{3 + \sqrt{13}}{2}] \cup [- \frac{3 - \sqrt{13}}{2}, \infty)$

Step 12