Convert to Set Notation x^2-8x+15<0

$x^{2} - 8 x + 15 < 0$

Step 1

Solve $3 < x < 5$ .

Convert the inequality to an equation.

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

Factor $x^{2} - 8 x + 15$ 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 $15$ and whose sum is $- 8$ .

$- 5, - 3$

Write the factored form using these integers.

$(x - 5) (x - 3) = 0$

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

$x - 5 = 0$

$x - 3 = 0$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

The final solution is all the values that make $(x - 5) (x - 3) = 0$ true.

$x = 5, 3$

Use each root to create test intervals.

$x < 3$

$3 < x < 5$

$x > 5$

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 $x < 3$ 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.

$x = 0$

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

${(0)}^{2} - 8 \cdot 0 + 15 < 0$

The left side $15$ is not less than the right side $0$ , which means that the given statement is false.

False

Test a value on the interval $3 < x < 5$ to see if it makes the inequality true.

Choose a value on the interval $3 < x < 5$ and see if this value makes the original inequality true.

$x = 4$

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

${(4)}^{2} - 8 \cdot 4 + 15 < 0$

The left side $- 1$ is less than the right side $0$ , which means that the given statement is always true.

True

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.

${(8)}^{2} - 8 \cdot 8 + 15 < 0$

The left side $15$ is not less than the right side $0$ , which means that the given statement is false.

False

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

$x < 3$ False

$3 < x < 5$ True

$x > 5$ False

$x < 3$ False

$3 < x < 5$ True

$x > 5$ False

The solution consists of all of the true intervals.

$3 < x < 5$

Step 2

Use the inequality $3 < x < 5$ to build the set notation.

${x | 3 < x < 5}$

Step 3

Convert to Set Notation x^2-8x+15<0

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