Solve the Inequality for x 2x^2-5x+15>3x+9

$2 x^{2} - 5 x + 15 > 3 x + 9$

Step 1

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

Subtract $3 x$ from both sides of the inequality.

$2 x^{2} - 5 x + 15 - 3 x > 9$

Subtract $3 x$ from $- 5 x$ .

$2 x^{2} - 8 x + 15 > 9$

Step 2

Convert the inequality to an equation.

$2 x^{2} - 8 x + 15 = 9$

Step 3

Subtract $9$ from both sides of the equation.

$2 x^{2} - 8 x + 15 - 9 = 0$

Step 4

Subtract $9$ from $15$ .

$2 x^{2} - 8 x + 6 = 0$

Step 5

Factor the left side of the equation.

Factor $2$ out of $2 x^{2} - 8 x + 6$ .

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

$2 (x^{2}) - 8 x + 6 = 0$

Factor $2$ out of $- 8 x$ .

$2 (x^{2}) + 2 (- 4 x) + 6 = 0$

Factor $2$ out of $6$ .

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

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

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

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

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

Factor.

Factor $x^{2} - 4 x + 3$ 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 $3$ and whose sum is $- 4$ .

$- 3, - 1$

Write the factored form using these integers.

$2 ((x - 3) (x - 1)) = 0$

Remove unnecessary parentheses.

$2 (x - 3) (x - 1) = 0$

Step 6

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

$x - 3 = 0$

$x - 1 = 0$

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 $x - 1$ 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 9

The final solution is all the values that make $2 (x - 3) (x - 1) = 0$ true.

$x = 3, 1$

Step 10

Use each root to create test intervals.

$x < 1$

$1 < x < 3$

$x > 3$

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 $x < 1$ 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.

$x = 0$

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

$2 {(0)}^{2} - 5 \cdot 0 + 15 > 3 (0) + 9$

The left side $15$ is greater than the right side $9$ , which means that the given statement is always true.

True

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

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

$x = 2$

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

$2 {(2)}^{2} - 5 \cdot 2 + 15 > 3 (2) + 9$

The left side $13$ is not greater than the right side $15$ , 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.

$2 {(6)}^{2} - 5 \cdot 6 + 15 > 3 (6) + 9$

The left side $57$ is greater than the right side $27$ , which means that the given statement is always true.

True

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

$x < 1$ True

$1 < x < 3$ False

$x > 3$ True

$x < 1$ True

$1 < x < 3$ False

$x > 3$ True

Step 12

The solution consists of all of the true intervals.

$x < 1$ or $x > 3$

Step 13

The result can be shown in multiple forms.

Inequality Form:

$x < 1 or x > 3$

Interval Notation:

$(- \infty, 1) \cup (3, \infty)$

Step 14

Solve the Inequality for x 2x^2-5x+15>3x+9

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