Solve by Factoring x^4-25=0

$x^{4} - 25 = 0$

Step 1

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

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

Step 2

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

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

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 $b = 5$ .

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

Step 4

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

$x^{2} + 5 = 0$

$x^{2} - 5 = 0$

Step 5

Set $x^{2} + 5$ equal to $0$ and solve for $x$ .

Set $x^{2} + 5$ equal to $0$ .

$x^{2} + 5 = 0$

Solve $x^{2} + 5 = 0$ for $x$ .

Subtract $5$ from both sides of the equation.

$x^{2} = - 5$

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

$x = \pm \sqrt{- 5}$

Simplify $\pm \sqrt{- 5}$ .

Rewrite $- 5$ as $- 1 (5)$ .

$x = \pm \sqrt{- 1 (5)}$

Rewrite $\sqrt{- 1 (5)}$ as $\sqrt{- 1} \cdot \sqrt{5}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{5}$

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

$x = \pm i \sqrt{5}$

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

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

$x = i \sqrt{5}$

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

$x = - i \sqrt{5}$

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

$x = i \sqrt{5}, - i \sqrt{5}$

Step 6

Set $x^{2} - 5$ equal to $0$ and solve for $x$ .

Set $x^{2} - 5$ equal to $0$ .

$x^{2} - 5 = 0$

Solve $x^{2} - 5 = 0$ for $x$ .

Add $5$ to both sides of the equation.

$x^{2} = 5$

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

$x = \pm \sqrt{5}$

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

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

$x = \sqrt{5}$

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

$x = - \sqrt{5}$

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

$x = \sqrt{5}, - \sqrt{5}$

Step 7

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

$x = i \sqrt{5}, - i \sqrt{5}, \sqrt{5}, - \sqrt{5}$

Solve by Factoring x^4-25=0

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