Solve the Inequality for x (x^2-3x-10)/(1-x)>=2 ?

$\frac{x^{2} - 3 x - 10}{1 - x} \geq 2$ ?

Subtract $2$ from both sides of the inequality.

$\frac{x^{2} - 3 x - 10}{1 - x} - 2 \geq 0$

Simplify $\frac{x^{2} - 3 x - 10}{1 - x} - 2$ .

Factor $x^{2} - 3 x - 10$ 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 $- 10$ and whose sum is $- 3$ .

$- 5, 2$

Write the factored form using these integers.

$\frac{(x - 5) (x + 2)}{1 - x} - 2 \geq 0$

To write $- 2$ as a fraction with a common denominator, multiply by $\frac{1 - x}{1 - x}$ .

$\frac{(x - 5) (x + 2)}{1 - x} - 2 \cdot \frac{1 - x}{1 - x} \geq 0$

Combine $- 2$ and $\frac{1 - x}{1 - x}$ .

$\frac{(x - 5) (x + 2)}{1 - x} + \frac{- 2 (1 - x)}{1 - x} \geq 0$

Combine the numerators over the common denominator.

$\frac{(x - 5) (x + 2) - 2 (1 - x)}{1 - x} \geq 0$

Simplify the numerator.

Expand $(x - 5) (x + 2)$ using the FOIL Method.

Apply the distributive property.

$\frac{x (x + 2) - 5 (x + 2) - 2 (1 - x)}{1 - x} \geq 0$

Apply the distributive property.

$\frac{x \cdot x + x \cdot 2 - 5 (x + 2) - 2 (1 - x)}{1 - x} \geq 0$

Apply the distributive property.

$\frac{x \cdot x + x \cdot 2 - 5 x - 5 \cdot 2 - 2 (1 - x)}{1 - x} \geq 0$

Simplify and combine like terms.

Simplify each term.

Multiply $x$ by $x$ .

$\frac{x^{2} + x \cdot 2 - 5 x - 5 \cdot 2 - 2 (1 - x)}{1 - x} \geq 0$

Move $2$ to the left of $x$ .

$\frac{x^{2} + 2 \cdot x - 5 x - 5 \cdot 2 - 2 (1 - x)}{1 - x} \geq 0$

Multiply $- 5$ by $2$ .

$\frac{x^{2} + 2 x - 5 x - 10 - 2 (1 - x)}{1 - x} \geq 0$

Subtract $5 x$ from $2 x$ .

$\frac{x^{2} - 3 x - 10 - 2 (1 - x)}{1 - x} \geq 0$

Apply the distributive property.

$\frac{x^{2} - 3 x - 10 - 2 \cdot 1 - 2 (- x)}{1 - x} \geq 0$

Multiply $- 2$ by $1$ .

$\frac{x^{2} - 3 x - 10 - 2 - 2 (- x)}{1 - x} \geq 0$

Multiply $- 1$ by $- 2$ .

$\frac{x^{2} - 3 x - 10 - 2 + 2 x}{1 - x} \geq 0$

Add $- 3 x$ and $2 x$ .

$\frac{x^{2} - x - 10 - 2}{1 - x} \geq 0$

Subtract $2$ from $- 10$ .

$\frac{x^{2} - x - 12}{1 - x} \geq 0$

Factor $x^{2} - x - 12$ 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 $- 12$ and whose sum is $- 1$ .

$- 4, 3$

Write the factored form using these integers.

$\frac{(x - 4) (x + 3)}{1 - x} \geq 0$

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

$x - 4 = 0$

$x + 3 = 0$

$1 - x = 0$

Add $4$ to both sides of the equation.

$x = 4$

Subtract $3$ from both sides of the equation.

$x = - 3$

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

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

$x = 4$

$x = - 3$

$x = 1$

Consolidate the solutions.

$x = 4, - 3, 1$

Find the domain of $\frac{(x - 4) (x + 3)}{1 - x}$ .

Set the denominator in $\frac{(x - 4) (x + 3)}{1 - x}$ equal to $0$ to find where the expression is undefined.

$1 - x = 0$

Solve for $x$ .

Subtract $1$ from both sides of the equation.

$- x = - 1$

Divide each term in $- x = - 1$ by $- 1$ and simplify.

Divide each term in $- x = - 1$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 1}{- 1}$

Simplify the left side.

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 1}{- 1}$

Divide $x$ by $1$ .

$x = \frac{- 1}{- 1}$

Simplify the right side.

Divide $- 1$ by $- 1$ .

$x = 1$

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

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

Use each root to create test intervals.

$x < - 3$

$- 3 < x < 1$

$1 < x < 4$

$x > 4$

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 $x < - 3$ to see if it makes the inequality true.

Choose a value on the interval $x < - 3$ and see if this value makes the original inequality true.

$x = - 6$

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

$\frac{{(- 6)}^{2} - 3 \cdot - 6 - 10}{1 - (- 6)} \geq 2$

The left side $6.\overline{285714}$ is greater than the right side $2$ , which means that the given statement is always true.

True

Test a value on the interval $- 3 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $- 3 < x < 1$ and see if this value makes the original inequality true.

$x = 0$

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

$\frac{{(0)}^{2} - 3 \cdot 0 - 10}{1 - (0)} \geq 2$

The left side $- 10$ is less than the right side $2$ , which means that the given statement is false.

False

Test a value on the interval $1 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 4$ and see if this value makes the original inequality true.

$x = 2$

Replace $x$ with $2$ in the original inequality.

$\frac{{(2)}^{2} - 3 \cdot 2 - 10}{1 - (2)} \geq 2$

The left side $12$ is greater than the right side $2$ , which means that the given statement is always true.

True

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.

$\frac{{(6)}^{2} - 3 \cdot 6 - 10}{1 - (6)} \geq 2$

The left side $- 1.6$ is less than the right side $2$ , which means that the given statement is false.

False

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

$x < - 3$ True

$- 3 < x < 1$ False

$1 < x < 4$ True

$x > 4$ False

$x < - 3$ True

$- 3 < x < 1$ False

$1 < x < 4$ True

$x > 4$ False

The solution consists of all of the true intervals.

$x \leq - 3$ or $1 < x \leq 4$

The result can be shown in multiple forms.

Inequality Form:

$x \leq - 3 or 1 < x \leq 4$

Interval Notation:

$(- \infty, - 3] \cup (1, 4]$

Solve the Inequality for x (x^2-3x-10)/(1-x)>=2 ?

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