Precalculus Examples

$1250 = \frac{1}{10} x (300 - x)$

Step 1

Rewrite the equation as $\frac{1}{10} \cdot (x (300 - x)) = 1250$ .

$\frac{1}{10} \cdot (x (300 - x)) = 1250$

Step 2

Multiply both sides of the equation by $10$ .

$10 (\frac{1}{10} \cdot (x (300 - x))) = 10 \cdot 1250$

Step 3

Simplify both sides of the equation.

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Step 3.1

Simplify the left side.

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Step 3.1.1

Simplify $10 (\frac{1}{10} \cdot (x (300 - x)))$ .

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Step 3.1.1.1

Simplify by multiplying through.

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Step 3.1.1.1.1

Apply the distributive property.

$10 (\frac{1}{10} \cdot (x \cdot 300 + x (- x))) = 10 \cdot 1250$

Step 3.1.1.1.2

Reorder.

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Step 3.1.1.1.2.1

Move $300$ to the left of $x$ .

$10 (\frac{1}{10} \cdot (300 \cdot x + x (- x))) = 10 \cdot 1250$

Step 3.1.1.1.2.2

Rewrite using the commutative property of multiplication.

$10 (\frac{1}{10} \cdot (300 \cdot x - x \cdot x)) = 10 \cdot 1250$

Step 3.1.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 3.1.1.2.1

Move $x$ .

$10 (\frac{1}{10} \cdot (300 x - (x \cdot x))) = 10 \cdot 1250$

Step 3.1.1.2.2

Multiply $x$ by $x$ .

$10 (\frac{1}{10} \cdot (300 x - x^{2})) = 10 \cdot 1250$

Step 3.1.1.3

Simplify terms.

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Step 3.1.1.3.1

Apply the distributive property.

$10 (\frac{1}{10} (300 x) + \frac{1}{10} (- x^{2})) = 10 \cdot 1250$

Step 3.1.1.3.2

Cancel the common factor of $10$ .

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Step 3.1.1.3.2.1

Factor $10$ out of $300 x$ .

$10 (\frac{1}{10} (10 (30 x)) + \frac{1}{10} (- x^{2})) = 10 \cdot 1250$

Step 3.1.1.3.2.2

Cancel the common factor.

$10 (\frac{1}{10} (10 (30 x)) + \frac{1}{10} (- x^{2})) = 10 \cdot 1250$

Step 3.1.1.3.2.3

Rewrite the expression.

$10 (30 x + \frac{1}{10} (- x^{2})) = 10 \cdot 1250$

Step 3.1.1.3.3

Combine $\frac{1}{10}$ and $x^{2}$ .

$10 (30 x - \frac{x^{2}}{10}) = 10 \cdot 1250$

Step 3.1.1.3.4

Apply the distributive property.

$10 (30 x) + 10 (- \frac{x^{2}}{10}) = 10 \cdot 1250$

Step 3.1.1.3.5

Multiply $30$ by $10$ .

$300 x + 10 (- \frac{x^{2}}{10}) = 10 \cdot 1250$

Step 3.1.1.3.6

Cancel the common factor of $10$ .

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Step 3.1.1.3.6.1

Move the leading negative in $- \frac{x^{2}}{10}$ into the numerator.

$300 x + 10 \frac{- x^{2}}{10} = 10 \cdot 1250$

Step 3.1.1.3.6.2

Cancel the common factor.

$300 x + 10 \frac{- x^{2}}{10} = 10 \cdot 1250$

Step 3.1.1.3.6.3

Rewrite the expression.

$300 x - x^{2} = 10 \cdot 1250$

Step 3.2

Simplify the right side.

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Step 3.2.1

Multiply $10$ by $1250$ .

$300 x - x^{2} = 12500$

Step 4

Subtract $12500$ from both sides of the equation.

$300 x - x^{2} - 12500 = 0$

Step 5

Factor the left side of the equation.

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Step 5.1

Factor $- 1$ out of $300 x - x^{2} - 12500$ .

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Step 5.1.1

Reorder $300 x$ and $- x^{2}$ .

$- x^{2} + 300 x - 12500 = 0$

Step 5.1.2

Factor $- 1$ out of $- x^{2}$ .

$- (x^{2}) + 300 x - 12500 = 0$

Step 5.1.3

Factor $- 1$ out of $300 x$ .

$- (x^{2}) - (- 300 x) - 12500 = 0$

Step 5.1.4

Rewrite $- 12500$ as $- 1 (12500)$ .

$- (x^{2}) - (- 300 x) - 1 \cdot 12500 = 0$

Step 5.1.5

Factor $- 1$ out of $- (x^{2}) - (- 300 x)$ .

$- (x^{2} - 300 x) - 1 \cdot 12500 = 0$

Step 5.1.6

Factor $- 1$ out of $- (x^{2} - 300 x) - 1 (12500)$ .

$- (x^{2} - 300 x + 12500) = 0$

Step 5.2

Factor.

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Step 5.2.1

Factor $x^{2} - 300 x + 12500$ using the AC method.

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Step 5.2.1.1

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 $12500$ and whose sum is $- 300$ .

$- 250, - 50$

Step 5.2.1.2

Write the factored form using these integers.

$- ((x - 250) (x - 50)) = 0$

Step 5.2.2

Remove unnecessary parentheses.

$- (x - 250) (x - 50) = 0$

Step 6

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

$x - 250 = 0$

$x - 50 = 0$

Step 7

Set $x - 250$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x - 250$ equal to $0$ .

$x - 250 = 0$

Step 7.2

Add $250$ to both sides of the equation.

$x = 250$

Step 8

Set $x - 50$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x - 50$ equal to $0$ .

$x - 50 = 0$

Step 8.2

Add $50$ to both sides of the equation.

$x = 50$

Step 9

The final solution is all the values that make $- (x - 250) (x - 50) = 0$ true.

$x = 250, 50$