Solve for x x^2-7x+8>-2

$x^{2} - 7 x + 8 > - 2$

Convert the inequality to an equation.

$x^{2} - 7 x + 8 = - 2$

Add $2$ to both sides of the equation.

$x^{2} - 7 x + 8 + 2 = 0$

Add $8$ and $2$ .

$x^{2} - 7 x + 10 = 0$

Factor $x^{2} - 7 x + 10$ 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 $10$ and whose sum is $- 7$ .

$- 5, - 2$

Write the factored form using these integers.

$(x - 5) (x - 2) = 0$

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

$x - 5 = 0$

$x - 2 = 0$

Set $x - 5$ equal to $0$ and solve for $x$ .

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Add $5$ to both sides of the equation.

$x = 5$

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.

$x = 2$

The final solution is all the values that make $(x - 5) (x - 2) = 0$ true.

$x = 5, 2$

Use each root to create test intervals.

$x < 2$

$2 < x < 5$

$x > 5$

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.

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

The left side $8$ is greater than the right side $- 2$ , which means that the given statement is always true.

True

Test a value on the interval $2 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $2 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

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

${(4)}^{2} - 7 \cdot 4 + 8 > - 2$

The left side $- 4$ is not greater than the right side $- 2$ , which means that the given statement is false.

False

Test a value on the interval $x > 5$ to see if it makes the inequality true.

Choose a value on the interval $x > 5$ and see if this value makes the original inequality true.

$x = 8$

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

${(8)}^{2} - 7 \cdot 8 + 8 > - 2$

The left side $16$ is greater than the right side $- 2$ , which means that the given statement is always true.

True

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

$x < 2$ True

$2 < x < 5$ False

$x > 5$ True

$x < 2$ True

$2 < x < 5$ False

$x > 5$ True

The solution consists of all of the true intervals.

$x < 2$ or $x > 5$

The result can be shown in multiple forms.

Inequality Form:

$x < 2 or x > 5$

Interval Notation:

$(- \infty, 2) \cup (5, \infty)$

Solve for x x^2-7x+8>-2

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