Solve the Inequality for x x^2-3(x+1)<=x-3x-3

$x^{2} - 3 (x + 1) \leq x - 3 x - 3$

Step 1

Simplify each term.

Apply the distributive property.

$x^{2} - 3 x - 3 \cdot 1 \leq x - 3 x - 3$

Multiply $- 3$ by $1$ .

$x^{2} - 3 x - 3 \leq x - 3 x - 3$

Step 2

Subtract $3 x$ from $x$ .

$x^{2} - 3 x - 3 \leq - 2 x - 3$

Step 3

Move all terms containing $x$ to the left side of the inequality.

Add $2 x$ to both sides of the inequality.

$x^{2} - 3 x - 3 + 2 x \leq - 3$

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

$x^{2} - x - 3 \leq - 3$

Step 4

Convert the inequality to an equation.

$x^{2} - x - 3 = - 3$

Step 5

Add $3$ to both sides of the equation.

$x^{2} - x - 3 + 3 = 0$

Step 6

Combine the opposite terms in $x^{2} - x - 3 + 3$ .

Add $- 3$ and $3$ .

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

Add $x^{2} - x$ and $0$ .

$x^{2} - x = 0$

Step 7

Factor $x$ out of $x^{2} - x$ .

Factor $x$ out of $x^{2}$ .

$x \cdot x - x = 0$

Factor $x$ out of $- x$ .

$x \cdot x + x \cdot - 1 = 0$

Factor $x$ out of $x \cdot x + x \cdot - 1$ .

$x (x - 1) = 0$

Step 8

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

$x = 0$

$x - 1 = 0$

Step 9

Set $x$ equal to $0$ .

$x = 0$

Step 10

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 11

The final solution is all the values that make $x (x - 1) = 0$ true.

$x = 0, 1$

Step 12

Use each root to create test intervals.

$x < 0$

$0 < x < 1$

$x > 1$

Step 13

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 $x < 0$ to see if it makes the inequality true.

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

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

${(- 2)}^{2} - 3 ((- 2) + 1) \leq (- 2) - 3 \cdot - 2 - 3$

The left side $7$ is greater than the right side $1$ , which means that the given statement is false.

False

Test a value on the interval $0 < x < 1$ to see if it makes the inequality true.

Choose a value on the interval $0 < x < 1$ and see if this value makes the original inequality true.

$x = 0.5$

Replace $x$ with $0.5$ in the original inequality.

${(0.5)}^{2} - 3 ((0.5) + 1) \leq (0.5) - 3 \cdot 0.5 - 3$

The left side $- 4.25$ is less than the right side $- 4$ , which means that the given statement is always true.

True

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

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

$x = 4$

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

${(4)}^{2} - 3 ((4) + 1) \leq (4) - 3 \cdot 4 - 3$

The left side $1$ is greater than the right side $- 11$ , which means that the given statement is false.

False

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

$x < 0$ False

$0 < x < 1$ True

$x > 1$ False

$x < 0$ False

$0 < x < 1$ True

$x > 1$ False

Step 14

The solution consists of all of the true intervals.

$0 \leq x \leq 1$

Step 15

The result can be shown in multiple forms.

Inequality Form:

$0 \leq x \leq 1$

Interval Notation:

$[0, 1]$

Step 16

Solve the Inequality for x x^2-3(x+1)<=x-3x-3

Related Questions

Support LCMGCF.com by disabling your adblocker.