Solve for x 0>x^2+5x-2

$0 > x^{2} + 5 x - 2$

Step 1

Rewrite so $x$ is on the left side of the inequality.

$x^{2} + 5 x - 2 < 0$

Step 2

Convert the inequality to an equation.

$x^{2} + 5 x - 2 = 0$

Step 3

Use the quadratic formula to find the solutions.

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

Step 4

Substitute the values $a = 1$ , $b = 5$ , and $c = - 2$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 2)}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

Raise $5$ to the power of $2$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 2$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 2}}{2 \cdot 1}$

Multiply $- 4$ by $- 2$ .

$x = \frac{- 5 \pm \sqrt{25 + 8}}{2 \cdot 1}$

Add $25$ and $8$ .

$x = \frac{- 5 \pm \sqrt{33}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm \sqrt{33}}{2}$

Step 6

Consolidate the solutions.

$x = - \frac{5 - \sqrt{33}}{2}, - \frac{5 + \sqrt{33}}{2}$

Step 7

Use each root to create test intervals.

$x < - \frac{5 + \sqrt{33}}{2}$

$- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$

$x > - \frac{5 - \sqrt{33}}{2}$

Step 8

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 $x < - \frac{5 + \sqrt{33}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{5 + \sqrt{33}}{2}$ and see if this value makes the original inequality true.

$x = - 8$

Replace $x$ with $- 8$ in the original inequality.

$0 > {(- 8)}^{2} + 5 (- 8) - 2$

The left side $0$ is not greater than the right side $22$ , which means that the given statement is false.

False

Test a value on the interval $- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

$0 > {(0)}^{2} + 5 (0) - 2$

The left side $0$ is greater than the right side $- 2$ , which means that the given statement is always true.

True

Test a value on the interval $x > - \frac{5 - \sqrt{33}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{5 - \sqrt{33}}{2}$ and see if this value makes the original inequality true.

$x = 3$

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

$0 > {(3)}^{2} + 5 (3) - 2$

The left side $0$ is not greater than the right side $22$ , which means that the given statement is false.

False

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

$x < - \frac{5 + \sqrt{33}}{2}$ False

$- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$ True

$x > - \frac{5 - \sqrt{33}}{2}$ False

$x < - \frac{5 + \sqrt{33}}{2}$ False

$- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$ True

$x > - \frac{5 - \sqrt{33}}{2}$ False

Step 9

The solution consists of all of the true intervals.

$- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$- \frac{5 + \sqrt{33}}{2} < x < - \frac{5 - \sqrt{33}}{2}$

Interval Notation:

$(- \frac{5 + \sqrt{33}}{2}, - \frac{5 - \sqrt{33}}{2})$

Step 11

Solve for x 0>x^2+5x-2

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