Pre-Algebra Examples

$\frac{1}{42^{2}} = \frac{1}{4 x} + \frac{5}{4 x^{2}}$

Step 1

Rewrite the equation as $\frac{1}{4 x} + \frac{5}{4 x^{2}} = \frac{1}{42^{2}}$ .

$\frac{1}{4 x} + \frac{5}{4 x^{2}} = \frac{1}{42^{2}}$

Step 2

Raise $42$ to the power of $2$ .

$\frac{1}{4 x} + \frac{5}{4 x^{2}} = \frac{1}{1764}$

Step 3

Find the LCD of the terms in the equation.

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Step 3.1

Finding the LCD of a list of values is the same as finding the LCM of the denominators of those values.

$4 x, 4 x^{2}, 1764$

Step 3.2

Since $4 x, 4 x^{2}, 1764$ contains both numbers and variables, there are two steps to find the LCM. Find LCM for the numeric part $4, 4, 1764$ then find LCM for the variable part $x^{1}, x^{2}$ .

Step 3.3

The LCM is the smallest positive number that all of the numbers divide into evenly.

1. List the prime factors of each number.

2. Multiply each factor the greatest number of times it occurs in either number.

Step 3.4

$4$ has factors of $2$ and $2$ .

$2 \cdot 2$

Step 3.5

The prime factors for $1764$ are $2 \cdot 2 \cdot 3 \cdot 3 \cdot 7 \cdot 7$ .

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Step 3.5.1

$1764$ has factors of $2$ and $882$ .

$2 \cdot 882$

Step 3.5.2

$882$ has factors of $2$ and $441$ .

$2 \cdot 2 \cdot 441$

Step 3.5.3

$441$ has factors of $3$ and $147$ .

$2 \cdot 2 \cdot 3 \cdot 147$

Step 3.5.4

$147$ has factors of $3$ and $49$ .

$2 \cdot 2 \cdot 3 \cdot 3 \cdot 49$

Step 3.5.5

$49$ has factors of $7$ and $7$ .

$2 \cdot 2 \cdot 3 \cdot 3 \cdot 7 \cdot 7$

Step 3.6

Multiply $2 \cdot 2 \cdot 3 \cdot 3 \cdot 7 \cdot 7$ .

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Step 3.6.1

Multiply $2$ by $2$ .

$4 \cdot 3 \cdot 3 \cdot 7 \cdot 7$

Step 3.6.2

Multiply $4$ by $3$ .

$12 \cdot 3 \cdot 7 \cdot 7$

Step 3.6.3

Multiply $12$ by $3$ .

$36 \cdot 7 \cdot 7$

Step 3.6.4

Multiply $36$ by $7$ .

$252 \cdot 7$

Step 3.6.5

Multiply $252$ by $7$ .

$1764$

Step 3.7

The factor for $x^{1}$ is $x$ itself.

$x^{1} = x$

$x$ occurs $1$ time.

Step 3.8

The factors for $x^{2}$ are $x \cdot x$ , which is $x$ multiplied by each other $2$ times.

$x^{2} = x \cdot x$

$x$ occurs $2$ times.

Step 3.9

The LCM of $x^{1}, x^{2}$ is the result of multiplying all prime factors the greatest number of times they occur in either term.

$x \cdot x$

Step 3.10

Multiply $x$ by $x$ .

$x^{2}$

Step 3.11

The LCM for $4 x, 4 x^{2}, 1764$ is the numeric part $1764$ multiplied by the variable part.

$1764 x^{2}$

Step 4

Multiply each term in $\frac{1}{4 x} + \frac{5}{4 x^{2}} = \frac{1}{1764}$ by $1764 x^{2}$ to eliminate the fractions.

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Step 4.1

Multiply each term in $\frac{1}{4 x} + \frac{5}{4 x^{2}} = \frac{1}{1764}$ by $1764 x^{2}$ .

$\frac{1}{4 x} (1764 x^{2}) + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify each term.

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Step 4.2.1.1

Rewrite using the commutative property of multiplication.

$1764 \frac{1}{4 x} x^{2} + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.2

Cancel the common factor of $4$ .

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Step 4.2.1.2.1

Factor $4$ out of $1764$ .

$4 (441) \frac{1}{4 x} x^{2} + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.2.2

Factor $4$ out of $4 x$ .

$4 (441) \frac{1}{4 (x)} x^{2} + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.2.3

Cancel the common factor.

$4 \cdot 441 \frac{1}{4 x} x^{2} + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.2.4

Rewrite the expression.

$441 \frac{1}{x} x^{2} + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.3

Combine $441$ and $\frac{1}{x}$ .

$\frac{441}{x} x^{2} + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.4

Cancel the common factor of $x$ .

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Step 4.2.1.4.1

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

$\frac{441}{x} (x \cdot x) + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.4.2

Cancel the common factor.

$\frac{441}{x} (x \cdot x) + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.4.3

Rewrite the expression.

$441 x + \frac{5}{4 x^{2}} (1764 x^{2}) = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.5

Rewrite using the commutative property of multiplication.

$441 x + 1764 \frac{5}{4 x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.6

Cancel the common factor of $4$ .

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Step 4.2.1.6.1

Factor $4$ out of $1764$ .

$441 x + 4 (441) \frac{5}{4 x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.6.2

Factor $4$ out of $4 x^{2}$ .

$441 x + 4 (441) \frac{5}{4 (x^{2})} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.6.3

Cancel the common factor.

$441 x + 4 \cdot 441 \frac{5}{4 x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.6.4

Rewrite the expression.

$441 x + 441 \frac{5}{x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.7

Combine $441$ and $\frac{5}{x^{2}}$ .

$441 x + \frac{441 \cdot 5}{x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.8

Multiply $441$ by $5$ .

$441 x + \frac{2205}{x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.9

Cancel the common factor of $x^{2}$ .

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Step 4.2.1.9.1

Cancel the common factor.

$441 x + \frac{2205}{x^{2}} x^{2} = \frac{1}{1764} (1764 x^{2})$

Step 4.2.1.9.2

Rewrite the expression.

$441 x + 2205 = \frac{1}{1764} (1764 x^{2})$

Step 4.3

Simplify the right side.

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Step 4.3.1

Cancel the common factor of $1764$ .

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Step 4.3.1.1

Factor $1764$ out of $1764 x^{2}$ .

$441 x + 2205 = \frac{1}{1764} (1764 (x^{2}))$

Step 4.3.1.2

Cancel the common factor.

$441 x + 2205 = \frac{1}{1764} (1764 x^{2})$

Step 4.3.1.3

Rewrite the expression.

$441 x + 2205 = x^{2}$

Step 5

Solve the equation.

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Step 5.1

Subtract $x^{2}$ from both sides of the equation.

$441 x + 2205 - x^{2} = 0$

Step 5.2

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Step 5.3

Substitute the values $a = - 1$ , $b = 441$ , and $c = 2205$ into the quadratic formula and solve for $x$ .

$\frac{- 441 \pm \sqrt{441^{2} - 4 \cdot (- 1 \cdot 2205)}}{2 \cdot - 1}$

Step 5.4

Simplify.

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Step 5.4.1

Simplify the numerator.

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Step 5.4.1.1

Raise $441$ to the power of $2$ .

$x = \frac{- 441 \pm \sqrt{194481 - 4 \cdot - 1 \cdot 2205}}{2 \cdot - 1}$

Step 5.4.1.2

Multiply $- 4 \cdot - 1 \cdot 2205$ .

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Step 5.4.1.2.1

Multiply $- 4$ by $- 1$ .

$x = \frac{- 441 \pm \sqrt{194481 + 4 \cdot 2205}}{2 \cdot - 1}$

Step 5.4.1.2.2

Multiply $4$ by $2205$ .

$x = \frac{- 441 \pm \sqrt{194481 + 8820}}{2 \cdot - 1}$

Step 5.4.1.3

Add $194481$ and $8820$ .

$x = \frac{- 441 \pm \sqrt{203301}}{2 \cdot - 1}$

Step 5.4.1.4

Rewrite $203301$ as $21^{2} \cdot 461$ .

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Step 5.4.1.4.1

Factor $441$ out of $203301$ .

$x = \frac{- 441 \pm \sqrt{441 (461)}}{2 \cdot - 1}$

Step 5.4.1.4.2

Rewrite $441$ as $21^{2}$ .

$x = \frac{- 441 \pm \sqrt{21^{2} \cdot 461}}{2 \cdot - 1}$

Step 5.4.1.5

Pull terms out from under the radical.

$x = \frac{- 441 \pm 21 \sqrt{461}}{2 \cdot - 1}$

Step 5.4.2

Multiply $2$ by $- 1$ .

$x = \frac{- 441 \pm 21 \sqrt{461}}{- 2}$

Step 5.4.3

Simplify $\frac{- 441 \pm 21 \sqrt{461}}{- 2}$ .

$x = \frac{441 \pm 21 \sqrt{461}}{2}$

Step 5.5

The final answer is the combination of both solutions.

$x = \frac{441 + 21 \sqrt{461}}{2}, \frac{441 - 21 \sqrt{461}}{2}$

Step 6

The result can be shown in multiple forms.

Exact Form:

$x = \frac{441 + 21 \sqrt{461}}{2}, \frac{441 - 21 \sqrt{461}}{2}$

Decimal Form:

$x = 445.94456081 \dots, - 4.94456081 \dots$