Solve the Inequality for x 3x^2+9x-61>-3x+2

$3 x^{2} + 9 x - 61 > - 3 x + 2$

Step 1

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

Add $3 x$ to both sides of the inequality.

$3 x^{2} + 9 x - 61 + 3 x > 2$

Add $9 x$ and $3 x$ .

$3 x^{2} + 12 x - 61 > 2$

Step 2

Convert the inequality to an equation.

$3 x^{2} + 12 x - 61 = 2$

Step 3

Subtract $2$ from both sides of the equation.

$3 x^{2} + 12 x - 61 - 2 = 0$

Step 4

Subtract $2$ from $- 61$ .

$3 x^{2} + 12 x - 63 = 0$

Step 5

Factor the left side of the equation.

Factor $3$ out of $3 x^{2} + 12 x - 63$ .

Factor $3$ out of $3 x^{2}$ .

$3 (x^{2}) + 12 x - 63 = 0$

Factor $3$ out of $12 x$ .

$3 (x^{2}) + 3 (4 x) - 63 = 0$

Factor $3$ out of $- 63$ .

$3 x^{2} + 3 (4 x) + 3 \cdot - 21 = 0$

Factor $3$ out of $3 x^{2} + 3 (4 x)$ .

$3 (x^{2} + 4 x) + 3 \cdot - 21 = 0$

Factor $3$ out of $3 (x^{2} + 4 x) + 3 \cdot - 21$ .

$3 (x^{2} + 4 x - 21) = 0$

Factor.

Factor $x^{2} + 4 x - 21$ 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 $- 21$ and whose sum is $4$ .

$- 3, 7$

Write the factored form using these integers.

$3 ((x - 3) (x + 7)) = 0$

Remove unnecessary parentheses.

$3 (x - 3) (x + 7) = 0$

Step 6

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

$x - 3 = 0$

$x + 7 = 0$

Step 7

Set $x - 3$ equal to $0$ and solve for $x$ .

Set $x - 3$ equal to $0$ .

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Step 8

Set $x + 7$ equal to $0$ and solve for $x$ .

Set $x + 7$ equal to $0$ .

$x + 7 = 0$

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 9

The final solution is all the values that make $3 (x - 3) (x + 7) = 0$ true.

$x = 3, - 7$

Step 10

Use each root to create test intervals.

$x < - 7$

$- 7 < x < 3$

$x > 3$

Step 11

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 $x < - 7$ to see if it makes the inequality true.

Choose a value on the interval $x < - 7$ and see if this value makes the original inequality true.

$x = - 10$

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

$3 {(- 10)}^{2} + 9 (- 10) - 61 > - 3 \cdot - 10 + 2$

The left side $149$ is greater than the right side $32$ , which means that the given statement is always true.

True

Test a value on the interval $- 7 < x < 3$ to see if it makes the inequality true.

Choose a value on the interval $- 7 < x < 3$ and see if this value makes the original inequality true.

$x = 0$

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

$3 {(0)}^{2} + 9 (0) - 61 > - 3 \cdot 0 + 2$

The left side $- 61$ is not greater than the right side $2$ , which means that the given statement is false.

False

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.

$3 {(6)}^{2} + 9 (6) - 61 > - 3 \cdot 6 + 2$

The left side $101$ is greater than the right side $- 16$ , which means that the given statement is always true.

True

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

$x < - 7$ True

$- 7 < x < 3$ False

$x > 3$ True

$x < - 7$ True

$- 7 < x < 3$ False

$x > 3$ True

Step 12

The solution consists of all of the true intervals.

$x < - 7$ or $x > 3$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < - 7 or x > 3$

Interval Notation:

$(- \infty, - 7) \cup (3, \infty)$

Step 14

Solve the Inequality for x 3x^2+9x-61>-3x+2

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