Precalculus Examples

$49 x = x^{3}$

Step 1

Subtract $x^{3}$ from both sides of the equation.

$49 x - x^{3} = 0$

Step 2

Factor the left side of the equation.

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Step 2.1

Factor $x$ out of $49 x - x^{3}$ .

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Step 2.1.1

Factor $x$ out of $49 x$ .

$x \cdot 49 - x^{3} = 0$

Step 2.1.2

Factor $x$ out of $- x^{3}$ .

$x \cdot 49 + x (- x^{2}) = 0$

Step 2.1.3

Factor $x$ out of $x \cdot 49 + x (- x^{2})$ .

$x (49 - x^{2}) = 0$

Step 2.2

Rewrite $49$ as $7^{2}$ .

$x (7^{2} - x^{2}) = 0$

Step 2.3

Factor.

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Step 2.3.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 7$ and $b = x$ .

$x ((7 + x) (7 - x)) = 0$

Step 2.3.2

Remove unnecessary parentheses.

$x (7 + x) (7 - x) = 0$

Step 3

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

$x = 0$

$7 + x = 0$

$7 - x = 0$

Step 4

Set $x$ equal to $0$ .

$x = 0$

Step 5

Set $7 + x$ equal to $0$ and solve for $x$ .

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Step 5.1

Set $7 + x$ equal to $0$ .

$7 + x = 0$

Step 5.2

Subtract $7$ from both sides of the equation.

$x = - 7$

Step 6

Set $7 - x$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $7 - x$ equal to $0$ .

$7 - x = 0$

Step 6.2

Solve $7 - x = 0$ for $x$ .

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Step 6.2.1

Subtract $7$ from both sides of the equation.

$- x = - 7$

Step 6.2.2

Divide each term in $- x = - 7$ by $- 1$ and simplify.

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Step 6.2.2.1

Divide each term in $- x = - 7$ by $- 1$ .

$\frac{- x}{- 1} = \frac{- 7}{- 1}$

Step 6.2.2.2

Simplify the left side.

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Step 6.2.2.2.1

Dividing two negative values results in a positive value.

$\frac{x}{1} = \frac{- 7}{- 1}$

Step 6.2.2.2.2

Divide $x$ by $1$ .

$x = \frac{- 7}{- 1}$

Step 6.2.2.3

Simplify the right side.

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Step 6.2.2.3.1

Divide $- 7$ by $- 1$ .

$x = 7$

Step 7

The final solution is all the values that make $x (7 + x) (7 - x) = 0$ true.

$x = 0, - 7, 7$