Algebra Examples

$x^{4} - 625 = 0$

Step 1

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

${(x^{2})}^{2} - 625 = 0$

Step 2

Rewrite $625$ as $25^{2}$ .

${(x^{2})}^{2} - 25^{2} = 0$

Step 3

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x^{2}$ and $b = 25$ .

$(x^{2} + 25) (x^{2} - 25) = 0$

Step 4

Simplify.

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Step 4.1

Rewrite $25$ as $5^{2}$ .

$(x^{2} + 25) (x^{2} - 5^{2}) = 0$

Step 4.2

Factor.

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Step 4.2.1

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 5$ .

$(x^{2} + 25) ((x + 5) (x - 5)) = 0$

Step 4.2.2

Remove unnecessary parentheses.

$(x^{2} + 25) (x + 5) (x - 5) = 0$

Step 5

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

$x^{2} + 25 = 0$

$x + 5 = 0$

$x - 5 = 0$

Step 6

Set $x^{2} + 25$ equal to $0$ and solve for $x$ .

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Step 6.1

Set $x^{2} + 25$ equal to $0$ .

$x^{2} + 25 = 0$

Step 6.2

Solve $x^{2} + 25 = 0$ for $x$ .

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Step 6.2.1

Subtract $25$ from both sides of the equation.

$x^{2} = - 25$

Step 6.2.2

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

$x = \pm \sqrt{- 25}$

Step 6.2.3

Simplify $\pm \sqrt{- 25}$ .

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Step 6.2.3.1

Rewrite $- 25$ as $- 1 (25)$ .

$x = \pm \sqrt{- 1 (25)}$

Step 6.2.3.2

Rewrite $\sqrt{- 1 (25)}$ as $\sqrt{- 1} \cdot \sqrt{25}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{25}$

Step 6.2.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{25}$

Step 6.2.3.4

Rewrite $25$ as $5^{2}$ .

$x = \pm i \cdot \sqrt{5^{2}}$

Step 6.2.3.5

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

$x = \pm i \cdot 5$

Step 6.2.3.6

Move $5$ to the left of $i$ .

$x = \pm 5 i$

Step 6.2.4

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

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Step 6.2.4.1

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

$x = 5 i$

Step 6.2.4.2

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

$x = - 5 i$

Step 6.2.4.3

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

$x = 5 i, - 5 i$

Step 7

Set $x + 5$ equal to $0$ and solve for $x$ .

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Step 7.1

Set $x + 5$ equal to $0$ .

$x + 5 = 0$

Step 7.2

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 8

Set $x - 5$ equal to $0$ and solve for $x$ .

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Step 8.1

Set $x - 5$ equal to $0$ .

$x - 5 = 0$

Step 8.2

Add $5$ to both sides of the equation.

$x = 5$

Step 9

The final solution is all the values that make $(x^{2} + 25) (x + 5) (x - 5) = 0$ true.

$x = 5 i, - 5 i, - 5, 5$