Graph x^2+x-10>10

$x^{2} + x - 10 > 10$

Convert the inequality to an equation.

$x^{2} + x - 10 = 10$

Subtract $10$ from both sides of the equation.

$x^{2} + x - 10 - 10 = 0$

Subtract $10$ from $- 10$ .

$x^{2} + x - 20 = 0$

Factor $x^{2} + x - 20$ using the AC method.

Consider the form $x^{2} + b x + c$ . Find a pair of integers whose product is $c$ and whose sum is $b$ . In this case, whose product is $- 20$ and whose sum is $1$ .

$- 4, 5$

Write the factored form using these integers.

$(x - 4) (x + 5) = 0$

If any individual factor on the left side of the equation is equal to $0$ , the entire expression will be equal to $0$ .

$x - 4 = 0$

$x + 5 = 0$

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Set $x + 5$ equal to $0$ and solve for $x$ .

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Subtract $5$ from both sides of the equation.

$x = - 5$

The final solution is all the values that make $(x - 4) (x + 5) = 0$ true.

$x = 4, - 5$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 4$

$x > 4$

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 $x < - 5$ to see if it makes the inequality true.

Choose a value on the interval $x < - 5$ and see if this value makes the original inequality true.

$x = - 8$

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

${(- 8)}^{2} - 8 - 10 > 10$

The left side $46$ is greater than the right side $10$ , which means that the given statement is always true.

True

Test a value on the interval $- 5 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 4$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 0 - 10 > 10$

The left side $- 10$ is not greater than the right side $10$ , which means that the given statement is false.

False

Test a value on the interval $x > 4$ to see if it makes the inequality true.

Choose a value on the interval $x > 4$ and see if this value makes the original inequality true.

$x = 6$

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

${(6)}^{2} + 6 - 10 > 10$

The left side $32$ is greater than the right side $10$ , which means that the given statement is always true.

True

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

$x < - 5$ True

$- 5 < x < 4$ False

$x > 4$ True

$x < - 5$ True

$- 5 < x < 4$ False

$x > 4$ True

The solution consists of all of the true intervals.

$x < - 5$ or $x > 4$

Graph x^2+x-10>10

Related Questions

Support LCMGCF.com by disabling your adblocker.