Solve for x 5x+x^2>100

$5 x + x^{2} > 100$

Step 1

Subtract $100$ from both sides of the inequality.

$5 x + x^{2} - 100 > 0$

Step 2

Convert the inequality to an equation.

$5 x + x^{2} - 100 = 0$

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 $c = - 100$ into the quadratic formula and solve for $x$ .

$\frac{- 5 \pm \sqrt{5^{2} - 4 \cdot (1 \cdot - 100)}}{2 \cdot 1}$

Step 5

Simplify.

Simplify the numerator.

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

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot 1 \cdot - 100}}{2 \cdot 1}$

Multiply $- 4 \cdot 1 \cdot - 100$ .

Multiply $- 4$ by $1$ .

$x = \frac{- 5 \pm \sqrt{25 - 4 \cdot - 100}}{2 \cdot 1}$

Multiply $- 4$ by $- 100$ .

$x = \frac{- 5 \pm \sqrt{25 + 400}}{2 \cdot 1}$

Add $25$ and $400$ .

$x = \frac{- 5 \pm \sqrt{425}}{2 \cdot 1}$

Rewrite $425$ as $5^{2} \cdot 17$ .

Factor $25$ out of $425$ .

$x = \frac{- 5 \pm \sqrt{25 (17)}}{2 \cdot 1}$

Rewrite $25$ as $5^{2}$ .

$x = \frac{- 5 \pm \sqrt{5^{2} \cdot 17}}{2 \cdot 1}$

Pull terms out from under the radical.

$x = \frac{- 5 \pm 5 \sqrt{17}}{2 \cdot 1}$

Multiply $2$ by $1$ .

$x = \frac{- 5 \pm 5 \sqrt{17}}{2}$

Step 6

Consolidate the solutions.

$x = - \frac{5 - 5 \sqrt{17}}{2}, - \frac{5 + 5 \sqrt{17}}{2}$

Step 7

Use each root to create test intervals.

$x < - \frac{5 + 5 \sqrt{17}}{2}$

$- \frac{5 + 5 \sqrt{17}}{2} < x < - \frac{5 - 5 \sqrt{17}}{2}$

$x > - \frac{5 - 5 \sqrt{17}}{2}$

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 $x < - \frac{5 + 5 \sqrt{17}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x < - \frac{5 + 5 \sqrt{17}}{2}$ and see if this value makes the original inequality true.

$x = - 15$

Replace $x$ with $- 15$ in the original inequality.

$5 (- 15) + {(- 15)}^{2} > 100$

The left side $150$ is greater than the right side $100$ , which means that the given statement is always true.

True

Test a value on the interval $- \frac{5 + 5 \sqrt{17}}{2} < x < - \frac{5 - 5 \sqrt{17}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $- \frac{5 + 5 \sqrt{17}}{2} < x < - \frac{5 - 5 \sqrt{17}}{2}$ and see if this value makes the original inequality true.

$x = 0$

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

$5 (0) + {(0)}^{2} > 100$

The left side $0$ is not greater than the right side $100$ , which means that the given statement is false.

False

Test a value on the interval $x > - \frac{5 - 5 \sqrt{17}}{2}$ to see if it makes the inequality true.

Choose a value on the interval $x > - \frac{5 - 5 \sqrt{17}}{2}$ and see if this value makes the original inequality true.

$x = 10$

Replace $x$ with $10$ in the original inequality.

$5 (10) + {(10)}^{2} > 100$

The left side $150$ is greater than the right side $100$ , which means that the given statement is always true.

True

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

$x < - \frac{5 + 5 \sqrt{17}}{2}$ True

$- \frac{5 + 5 \sqrt{17}}{2} < x < - \frac{5 - 5 \sqrt{17}}{2}$ False

$x > - \frac{5 - 5 \sqrt{17}}{2}$ True

$x < - \frac{5 + 5 \sqrt{17}}{2}$ True

$- \frac{5 + 5 \sqrt{17}}{2} < x < - \frac{5 - 5 \sqrt{17}}{2}$ False

$x > - \frac{5 - 5 \sqrt{17}}{2}$ True

Step 9

The solution consists of all of the true intervals.

$x < - \frac{5 + 5 \sqrt{17}}{2}$ or $x > - \frac{5 - 5 \sqrt{17}}{2}$

Step 10

The result can be shown in multiple forms.

Inequality Form:

$x < - \frac{5 + 5 \sqrt{17}}{2} or x > - \frac{5 - 5 \sqrt{17}}{2}$

Interval Notation:

$(- \infty, - \frac{5 + 5 \sqrt{17}}{2}) \cup (- \frac{5 - 5 \sqrt{17}}{2}, \infty)$

Step 11

Solve for x 5x+x^2>100

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