Solve the Inequality for x x^2+x-1>0

$x^{2} + x - 1 > 0$

Convert the inequality to an equation.

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

Use the quadratic formula to find the solutions.

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

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

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

Simplify.

Simplify the numerator.

One to any power is one.

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

Multiply $- 4$ by $1$ .

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

Multiply $- 4$ by $- 1$ .

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

Add $1$ and $4$ .

$x = \frac{- 1 \pm \sqrt{5}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 1 \pm \sqrt{5}}{2}$

Consolidate the solutions.

$x = - \frac{1 - \sqrt{5}}{2}, - \frac{1 + \sqrt{5}}{2}$

Use each root to create test intervals.

$x < - \frac{1 + \sqrt{5}}{2}$

$- \frac{1 + \sqrt{5}}{2} < x < - \frac{1 - \sqrt{5}}{2}$

$x > - \frac{1 - \sqrt{5}}{2}$

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

Test a value on the interval $x < - \frac{1 + \sqrt{5}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{1 + \sqrt{5}}{2}$ and see if this value makes the original inequality true.

$x = - 4$

Replace $x$ with $- 4$ in the original inequality.

${(- 4)}^{2} - 4 - 1 > 0$

The left side $11$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{1 + \sqrt{5}}{2} < x < - \frac{1 - \sqrt{5}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{1 + \sqrt{5}}{2} < x < - \frac{1 - \sqrt{5}}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 0 - 1 > 0$

The left side $- 1$ is not greater than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > - \frac{1 - \sqrt{5}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{1 - \sqrt{5}}{2}$ and see if this value makes the original inequality true.

$x = 3$

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

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

The left side $11$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < - \frac{1 + \sqrt{5}}{2}$ True

$- \frac{1 + \sqrt{5}}{2} < x < - \frac{1 - \sqrt{5}}{2}$ False

$x > - \frac{1 - \sqrt{5}}{2}$ True

$x < - \frac{1 + \sqrt{5}}{2}$ True

$- \frac{1 + \sqrt{5}}{2} < x < - \frac{1 - \sqrt{5}}{2}$ False

$x > - \frac{1 - \sqrt{5}}{2}$ True

The solution consists of all of the true intervals.

$x < - \frac{1 + \sqrt{5}}{2}$ or $x > - \frac{1 - \sqrt{5}}{2}$

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{1 + \sqrt{5}}{2} or x > - \frac{1 - \sqrt{5}}{2}$

Interval Notation:

$(- \infty, - \frac{1 + \sqrt{5}}{2}) \cup (- \frac{1 - \sqrt{5}}{2}, \infty)$

Solve the Inequality for x x^2+x-1>0

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