Finite Math Examples

$x^{2} - x = 0$

Step 1

To create a trinomial square on the left side of the equation, find a value that is equal to the square of half of $b$ .

${(\frac{b}{2})}^{2} = {(- \frac{1}{2})}^{2}$

Step 2

Add the term to each side of the equation.

$x^{2} - x + {(- \frac{1}{2})}^{2} = 0 + {(- \frac{1}{2})}^{2}$

Step 3

Simplify the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify each term.

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Step 3.1.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.1.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

$x^{2} - x + {(- 1)}^{2} {(\frac{1}{2})}^{2} = 0 + {(- \frac{1}{2})}^{2}$

Step 3.1.1.1.2

Apply the product rule to $\frac{1}{2}$ .

$x^{2} - x + {(- 1)}^{2} \frac{1^{2}}{2^{2}} = 0 + {(- \frac{1}{2})}^{2}$

Step 3.1.1.2

Raise $- 1$ to the power of $2$ .

$x^{2} - x + 1 \frac{1^{2}}{2^{2}} = 0 + {(- \frac{1}{2})}^{2}$

Step 3.1.1.3

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

$x^{2} - x + \frac{1^{2}}{2^{2}} = 0 + {(- \frac{1}{2})}^{2}$

Step 3.1.1.4

One to any power is one.

$x^{2} - x + \frac{1}{2^{2}} = 0 + {(- \frac{1}{2})}^{2}$

Step 3.1.1.5

Raise $2$ to the power of $2$ .

$x^{2} - x + \frac{1}{4} = 0 + {(- \frac{1}{2})}^{2}$

Step 3.2

Simplify the right side.

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Step 3.2.1

Simplify $0 + {(- \frac{1}{2})}^{2}$ .

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Step 3.2.1.1

Simplify each term.

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Step 3.2.1.1.1

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 3.2.1.1.1.1

Apply the product rule to $- \frac{1}{2}$ .

$x^{2} - x + \frac{1}{4} = 0 + {(- 1)}^{2} {(\frac{1}{2})}^{2}$

Step 3.2.1.1.1.2

Apply the product rule to $\frac{1}{2}$ .

$x^{2} - x + \frac{1}{4} = 0 + {(- 1)}^{2} \frac{1^{2}}{2^{2}}$

Step 3.2.1.1.2

Raise $- 1$ to the power of $2$ .

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

Step 3.2.1.1.3

Multiply $\frac{1^{2}}{2^{2}}$ by $1$ .

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

Step 3.2.1.1.4

One to any power is one.

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

Step 3.2.1.1.5

Raise $2$ to the power of $2$ .

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

Step 3.2.1.2

Add $0$ and $\frac{1}{4}$ .

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

Step 4

Factor the perfect trinomial square into ${(x - \frac{1}{2})}^{2}$ .

${(x - \frac{1}{2})}^{2} = \frac{1}{4}$

Step 5

Solve the equation for $x$ .

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Step 5.1

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x - \frac{1}{2} = \pm \sqrt{\frac{1}{4}}$

Step 5.2

Simplify $\pm \sqrt{\frac{1}{4}}$ .

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Step 5.2.1

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

$x - \frac{1}{2} = \pm \frac{\sqrt{1}}{\sqrt{4}}$

Step 5.2.2

Any root of $1$ is $1$ .

$x - \frac{1}{2} = \pm \frac{1}{\sqrt{4}}$

Step 5.2.3

Simplify the denominator.

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Step 5.2.3.1

Rewrite $4$ as $2^{2}$ .

$x - \frac{1}{2} = \pm \frac{1}{\sqrt{2^{2}}}$

Step 5.2.3.2

Pull terms out from under the radical, assuming positive real numbers.

$x - \frac{1}{2} = \pm \frac{1}{2}$

Step 5.3

The complete solution is the result of both the positive and negative portions of the solution.

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Step 5.3.1

First, use the positive value of the $\pm$ to find the first solution.

$x - \frac{1}{2} = \frac{1}{2}$

Step 5.3.2

Move all terms not containing $x$ to the right side of the equation.

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Step 5.3.2.1

Add $\frac{1}{2}$ to both sides of the equation.

$x = \frac{1}{2} + \frac{1}{2}$

Step 5.3.2.2

Combine the numerators over the common denominator.

$x = \frac{1 + 1}{2}$

Step 5.3.2.3

Add $1$ and $1$ .

$x = \frac{2}{2}$

Step 5.3.2.4

Divide $2$ by $2$ .

$x = 1$

Step 5.3.3

Next, use the negative value of the $\pm$ to find the second solution.

$x - \frac{1}{2} = - \frac{1}{2}$

Step 5.3.4

Move all terms not containing $x$ to the right side of the equation.

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Step 5.3.4.1

Add $\frac{1}{2}$ to both sides of the equation.

$x = - \frac{1}{2} + \frac{1}{2}$

Step 5.3.4.2

Combine the numerators over the common denominator.

$x = \frac{- 1 + 1}{2}$

Step 5.3.4.3

Add $- 1$ and $1$ .

$x = \frac{0}{2}$

Step 5.3.4.4

Divide $0$ by $2$ .

$x = 0$

Step 5.3.5

The complete solution is the result of both the positive and negative portions of the solution.

$x = 1, 0$