Algebra Examples

$- \frac{1}{4} x^{2} + 4 x + 1 \geq 1$

Step 1

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

$- \frac{x^{2}}{4} + 4 x + 1 \geq 1$

Step 2

Multiply each term in $- \frac{x^{2}}{4} + 4 x + 1 \geq 1$ by $4$ to eliminate the fractions.

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Step 2.1

Multiply each term in $- \frac{x^{2}}{4} + 4 x + 1 \geq 1$ by $4$ .

$- \frac{x^{2}}{4} \cdot 4 + 4 x \cdot 4 + 1 \cdot 4 \geq 1 \cdot 4$

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

Move the leading negative in $- \frac{x^{2}}{4}$ into the numerator.

$\frac{- x^{2}}{4} \cdot 4 + 4 x \cdot 4 + 1 \cdot 4 \geq 1 \cdot 4$

Step 2.2.1.1.2

Cancel the common factor.

$\frac{- x^{2}}{4} \cdot 4 + 4 x \cdot 4 + 1 \cdot 4 \geq 1 \cdot 4$

Step 2.2.1.1.3

Rewrite the expression.

$- x^{2} + 4 x \cdot 4 + 1 \cdot 4 \geq 1 \cdot 4$

Step 2.2.1.2

Multiply $4$ by $4$ .

$- x^{2} + 16 x + 1 \cdot 4 \geq 1 \cdot 4$

Step 2.2.1.3

Multiply $4$ by $1$ .

$- x^{2} + 16 x + 4 \geq 1 \cdot 4$

Step 2.3

Simplify the right side.

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Step 2.3.1

Multiply $4$ by $1$ .

$- x^{2} + 16 x + 4 \geq 4$

Step 3

Convert the inequality to an equation.

$- x^{2} + 16 x + 4 = 4$

Step 4

Subtract $4$ from both sides of the equation.

$- x^{2} + 16 x + 4 - 4 = 0$

Step 5

Combine the opposite terms in $- x^{2} + 16 x + 4 - 4$ .

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Step 5.1

Subtract $4$ from $4$ .

$- x^{2} + 16 x + 0 = 0$

Step 5.2

Add $- x^{2} + 16 x$ and $0$ .

$- x^{2} + 16 x = 0$

Step 6

Factor $- x$ out of $- x^{2} + 16 x$ .

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Step 6.1

Factor $- x$ out of $- x^{2}$ .

$- x \cdot x + 16 x = 0$

Step 6.2

Factor $- x$ out of $16 x$ .

$- x \cdot x - x \cdot - 16 = 0$

Step 6.3

Factor $- x$ out of $- x (x) - x (- 16)$ .

$- x (x - 16) = 0$

Step 7

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

$x = 0$

$x - 16 = 0$

Step 8

Set $x$ equal to $0$ .

$x = 0$

Step 9

Set $x - 16$ equal to $0$ and solve for $x$ .

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Step 9.1

Set $x - 16$ equal to $0$ .

$x - 16 = 0$

Step 9.2

Add $16$ to both sides of the equation.

$x = 16$

Step 10

The final solution is all the values that make $- x (x - 16) = 0$ true.

$x = 0, 16$

Step 11

Use each root to create test intervals.

$x < 0$

$0 < x < 16$

$x > 16$

Step 12

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 12.1

Test a value on the interval $x < 0$ to see if it makes the inequality true.

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Step 12.1.1

Choose a value on the interval $x < 0$ and see if this value makes the original inequality true.

$x = - 2$

Step 12.1.2

Replace $x$ with $- 2$ in the original inequality.

$- \frac{1}{4} \cdot {(- 2)}^{2} + 4 (- 2) + 1 \geq 1$

Step 12.1.3

The left side $- 8$ is less than the right side $1$ , which means that the given statement is false.

False

Step 12.2

Test a value on the interval $0 < x < 16$ to see if it makes the inequality true.

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Step 12.2.1

Choose a value on the interval $0 < x < 16$ and see if this value makes the original inequality true.

$x = 2$

Step 12.2.2

Replace $x$ with $2$ in the original inequality.

$- \frac{1}{4} \cdot {(2)}^{2} + 4 (2) + 1 \geq 1$

Step 12.2.3

The left side $8$ is greater than the right side $1$ , which means that the given statement is always true.

True

Step 12.3

Test a value on the interval $x > 16$ to see if it makes the inequality true.

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Step 12.3.1

Choose a value on the interval $x > 16$ and see if this value makes the original inequality true.

$x = 18$

Step 12.3.2

Replace $x$ with $18$ in the original inequality.

$- \frac{1}{4} \cdot {(18)}^{2} + 4 (18) + 1 \geq 1$

Step 12.3.3

The left side $- 8$ is less than the right side $1$ , which means that the given statement is false.

False

Step 12.4

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

$x < 0$ False

$0 < x < 16$ True

$x > 16$ False

$x < 0$ False

$0 < x < 16$ True

$x > 16$ False

Step 13

The solution consists of all of the true intervals.

$0 \leq x \leq 16$

Step 14

The result can be shown in multiple forms.

Inequality Form:

$0 \leq x \leq 16$

Interval Notation:

$[0, 16]$

Step 15