Algebra Examples

$k^{2} - 7 k + 12.25 > 0$

Step 1

Convert the inequality to an equation.

$k^{2} - 7 k + 12.25 = 0$

Step 2

Factor using the perfect square rule.

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Step 2.1

Rewrite $12.25$ as ${3.5}^{2}$ .

$k^{2} - 7 k + {3.5}^{2} = 0$

Step 2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$7 k = 2 \cdot k \cdot 3.5$

Step 2.3

Rewrite the polynomial.

$k^{2} - 2 \cdot k \cdot 3.5 + {3.5}^{2} = 0$

Step 2.4

Factor using the perfect square trinomial rule $a^{2} - 2 a b + b^{2} = {(a - b)}^{2}$ , where $a = k$ and $b = 3.5$ .

${(k - 3.5)}^{2} = 0$

Step 3

Set the $k - 3.5$ equal to $0$ .

$k - 3.5 = 0$

Step 4

Add $3.5$ to both sides of the equation.

$k = 3.5$

Step 5

Use each root to create test intervals.

$k < 3.5$

$k > 3.5$

Step 6

Choose a test value from each interval and plug this value into the original inequality to determine which intervals satisfy the inequality.

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Step 6.1

Test a value on the interval $k < 3.5$ to see if it makes the inequality true.

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Step 6.1.1

Choose a value on the interval $k < 3.5$ and see if this value makes the original inequality true.

$k = 0$

Step 6.1.2

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

${(0)}^{2} - 7 \cdot 0 + 12.25 > 0$

Step 6.1.3

The left side $12.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.2

Test a value on the interval $k > 3.5$ to see if it makes the inequality true.

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Step 6.2.1

Choose a value on the interval $k > 3.5$ and see if this value makes the original inequality true.

$k = 6$

Step 6.2.2

Replace $k$ with $6$ in the original inequality.

${(6)}^{2} - 7 \cdot 6 + 12.25 > 0$

Step 6.2.3

The left side $6.25$ is greater than the right side $0$ , which means that the given statement is always true.

True

Step 6.3

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

$k < 3.5$ True

$k > 3.5$ True

$k < 3.5$ True

$k > 3.5$ True

Step 7

The solution consists of all of the true intervals.

$k < 3.5$ or $k > 3.5$

Step 8

The result can be shown in multiple forms.

Inequality Form:

$k < 3.5 or k > 3.5$

Interval Notation:

$(- \infty, 3.5) \cup (3.5, \infty)$

Step 9