Finite Math Examples

$\frac{1}{4} x^{2} - 2 x + 3 \leq 0$

Step 1

Combine $\frac{1}{4}$ and $x^{2}$ .

$\frac{x^{2}}{4} - 2 x + 3 \leq 0$

Step 2

Multiply each term in $\frac{x^{2}}{4} - 2 x + 3 \leq 0$ by $4$ to eliminate the fractions.

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Step 2.1

Multiply each term in $\frac{x^{2}}{4} - 2 x + 3 \leq 0$ by $4$ .

$\frac{x^{2}}{4} \cdot 4 - 2 x \cdot 4 + 3 \cdot 4 \leq 0 \cdot 4$

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify each term.

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Step 2.2.1.1

Cancel the common factor of $4$ .

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Step 2.2.1.1.1

Cancel the common factor.

$\frac{x^{2}}{4} \cdot 4 - 2 x \cdot 4 + 3 \cdot 4 \leq 0 \cdot 4$

Step 2.2.1.1.2

Rewrite the expression.

$x^{2} - 2 x \cdot 4 + 3 \cdot 4 \leq 0 \cdot 4$

Step 2.2.1.2

Multiply $4$ by $- 2$ .

$x^{2} - 8 x + 3 \cdot 4 \leq 0 \cdot 4$

Step 2.2.1.3

Multiply $3$ by $4$ .

$x^{2} - 8 x + 12 \leq 0 \cdot 4$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply $0$ by $4$ .

$x^{2} - 8 x + 12 \leq 0$

Step 3

Convert the inequality to an equation.

$x^{2} - 8 x + 12 = 0$

Step 4

Factor $x^{2} - 8 x + 12$ using the AC method.

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Step 4.1

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 $12$ and whose sum is $- 8$ .

$- 6, - 2$

Step 4.2

Write the factored form using these integers.

$(x - 6) (x - 2) = 0$

Step 5

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

$x - 6 = 0$

$x - 2 = 0$

Step 6

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

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Step 6.1

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

$x - 6 = 0$

Step 6.2

Add $6$ to both sides of the equation.

$x = 6$

Step 7

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

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Step 7.1

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

$x - 2 = 0$

Step 7.2

Add $2$ to both sides of the equation.

$x = 2$

Step 8

The final solution is all the values that make $(x - 6) (x - 2) = 0$ true.

$x = 6, 2$

Step 9

Use each root to create test intervals.

$x < 2$

$2 < x < 6$

$x > 6$

Step 10

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 10.1

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

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Step 10.1.1

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

$x = 0$

Step 10.1.2

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

$\frac{1}{4} \cdot {(0)}^{2} - 2 \cdot 0 + 3 \leq 0$

Step 10.1.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.2

Test a value on the interval $2 < x < 6$ to see if it makes the inequality true.

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Step 10.2.1

Choose a value on the interval $2 < x < 6$ and see if this value makes the original inequality true.

$x = 4$

Step 10.2.2

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

$\frac{1}{4} \cdot {(4)}^{2} - 2 \cdot 4 + 3 \leq 0$

Step 10.2.3

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

True

Step 10.3

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

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Step 10.3.1

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

$x = 8$

Step 10.3.2

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

$\frac{1}{4} \cdot {(8)}^{2} - 2 \cdot 8 + 3 \leq 0$

Step 10.3.3

The left side $3$ is greater than the right side $0$ , which means that the given statement is false.

False

Step 10.4

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

$x < 2$ False

$2 < x < 6$ True

$x > 6$ False

$x < 2$ False

$2 < x < 6$ True

$x > 6$ False

Step 11

The solution consists of all of the true intervals.

$2 \leq x \leq 6$

Step 12

The result can be shown in multiple forms.

Inequality Form:

$2 \leq x \leq 6$

Interval Notation:

$[2, 6]$

Step 13