Solve the Inequality for x x^2-7x+6>=0

$x^{2} - 7 x + 6 \geq 0$

Convert the inequality to an equation.

$x^{2} - 7 x + 6 = 0$

Factor $x^{2} - 7 x + 6$ 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 $6$ and whose sum is $- 7$ .

$- 6, - 1$

Write the factored form using these integers.

$(x - 6) (x - 1) = 0$

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

$x - 6 = 0$

$x - 1 = 0$

Set $x - 6$ equal to $0$ and solve for $x$ .

Set $x - 6$ equal to $0$ .

$x - 6 = 0$

Add $6$ to both sides of the equation.

$x = 6$

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.

$x = 1$

The final solution is all the values that make $(x - 6) (x - 1) = 0$ true.

$x = 6, 1$

Use each root to create test intervals.

$x < 1$

$1 < x < 6$

$x > 6$

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.

${(0)}^{2} - 7 \cdot 0 + 6 \geq 0$

The left side $6$ is greater than the right side $0$ , which means that the given statement is always true.

True

Test a value on the interval $1 < x < 6$ to see if it makes the inequality true.

Choose a value on the interval $1 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

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

${(4)}^{2} - 7 \cdot 4 + 6 \geq 0$

The left side $- 6$ is less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $x > 6$ to see if it makes the inequality true.

Choose a value on the interval $x > 6$ and see if this value makes the original inequality true.

$x = 8$

Replace $x$ with $8$ in the original inequality.

${(8)}^{2} - 7 \cdot 8 + 6 \geq 0$

The left side $14$ is greater than the right side $0$ , which means that the given statement is always true.

True

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

$x < 1$ True

$1 < x < 6$ False

$x > 6$ True

$x < 1$ True

$1 < x < 6$ False

$x > 6$ True

The solution consists of all of the true intervals.

$x \leq 1$ or $x \geq 6$

The result can be shown in multiple forms.

Inequality Form:

$x \leq 1 or x \geq 6$

Interval Notation:

$(- \infty, 1] \cup [6, \infty)$

Solve the Inequality for x x^2-7x+6>=0

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