Solve the Inequality for x x^2+8>=6x

$x^{2} + 8 \geq 6 x$

Step 1

Subtract $6 x$ from both sides of the inequality.

$x^{2} + 8 - 6 x \geq 0$

Step 2

Convert the inequality to an equation.

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

Step 3

Factor $x^{2} + 8 - 6 x$ 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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Write the factored form using these integers.

$(x - 4) (x - 2) = 0$

Step 4

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

$x - 4 = 0$

$x - 2 = 0$

Step 5

Set $x - 4$ equal to $0$ and solve for $x$ .

Set $x - 4$ equal to $0$ .

$x - 4 = 0$

Add $4$ to both sides of the equation.

$x = 4$

Step 6

Set $x - 2$ equal to $0$ and solve for $x$ .

Set $x - 2$ equal to $0$ .

$x - 2 = 0$

Add $2$ to both sides of the equation.

$x = 2$

Step 7

The final solution is all the values that make $(x - 4) (x - 2) = 0$ true.

$x = 4, 2$

Step 8

Use each root to create test intervals.

$x < 2$

$2 < x < 4$

$x > 4$

Step 9

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 $x < 2$ to see if it makes the inequality true.

Choose a value on the interval $x < 2$ and see if this value makes the original inequality true.

$x = 0$

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

${(0)}^{2} + 8 \geq 6 (0)$

The left side $8$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $2 < x < 4$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 4$ and see if this value makes the original inequality true.

$x = 3$

Replace $x$ with $3$ in the original inequality.

${(3)}^{2} + 8 \geq 6 (3)$

The left side $17$ is less than the right side $18$ , which means that the given statement is false.

False

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.

${(6)}^{2} + 8 \geq 6 (6)$

The left side $44$ is greater than the right side $36$ , which means that the given statement is always true.

True

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

$x < 2$ True

$2 < x < 4$ False

$x > 4$ True

$x < 2$ True

$2 < x < 4$ False

$x > 4$ True

Step 10

The solution consists of all of the true intervals.

$x \leq 2$ or $x \geq 4$

Step 11

The result can be shown in multiple forms.

Inequality Form:

$x \leq 2 or x \geq 4$

Interval Notation:

$(- \infty, 2] \cup [4, \infty)$

Step 12

Solve the Inequality for x x^2+8>=6x

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