Solve for x (x^2-x-30)/(x-1)>=0

$\frac{x^{2} - x - 30}{x - 1} \geq 0$

Find all the values where the expression switches from negative to positive by setting each factor equal to $0$ and solving.

$x^{2} - x - 30 = 0$

$x - 1 = 0$

Factor $x^{2} - x - 30$ 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 $- 30$ and whose sum is $- 1$ .

$- 6, 5$

Write the factored form using these integers.

$(x - 6) (x + 5) = 0$

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

$x - 6 = 0$

$x + 5 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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 $(x - 6) (x + 5) = 0$ true.

$x = 6, - 5$

Add $1$ to both sides of the equation.

$x = 1$

Solve for each factor to find the values where the absolute value expression goes from negative to positive.

$x = 6, - 5$

$x = 1$

Consolidate the solutions.

$x = 6, - 5, 1$

Find the domain of $\frac{x^{2} - x - 30}{x - 1}$ .

Set the denominator in $\frac{x^{2} - x - 30}{x - 1}$ equal to $0$ to find where the expression is undefined.

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The domain is all values of $x$ that make the expression defined.

$(- \infty, 1) \cup (1, \infty)$

Use each root to create test intervals.

$x < - 5$

$- 5 < x < 1$

$1 < x < 6$

$x > 6$

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.

$\frac{{(- 8)}^{2} - (- 8) - 30}{(- 8) - 1} \geq 0$

The left side $- 4.\overline{6}$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $- 5 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 5 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - (0) - 30}{(0) - 1} \geq 0$

The left side $30$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

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

$\frac{{(4)}^{2} - (4) - 30}{(4) - 1} \geq 0$

The left side $- 6$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

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

$\frac{{(8)}^{2} - (8) - 30}{(8) - 1} \geq 0$

The left side $3.\overline{714285}$ 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.

$x < - 5$ False

$- 5 < x < 1$ True

$1 < x < 6$ False

$x > 6$ True

$x < - 5$ False

$- 5 < x < 1$ True

$1 < x < 6$ False

$x > 6$ True

The solution consists of all of the true intervals.

$- 5 \leq x < 1$ or $x \geq 6$

The result can be shown in multiple forms.

Inequality Form:

$- 5 \leq x < 1 or x \geq 6$

Interval Notation:

$[- 5, 1) \cup [6, \infty)$

Solve for x (x^2-x-30)/(x-1)>=0

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