Algebra Examples

$x^{4} - 30 x^{2} + 125 = 0$

Step 1

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

${(x^{2})}^{2} - 30 x^{2} + 125 = 0$

Step 2

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$u^{2} - 30 u + 125 = 0$

Step 3

Factor $u^{2} - 30 u + 125$ using the AC method.

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Step 3.1

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 $125$ and whose sum is $- 30$ .

$- 25, - 5$

Step 3.2

Write the factored form using these integers.

$(u - 25) (u - 5) = 0$

Step 4

Replace all occurrences of $u$ with $x^{2}$ .

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

Step 5

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

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

Step 6

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 $b = 5$ .

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

Step 7

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

$x + 5 = 0$

$x - 5 = 0$

$x^{2} - 5 = 0$

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

Subtract $5$ from both sides of the equation.

$x = - 5$

Step 9

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

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Step 9.1

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

$x - 5 = 0$

Step 9.2

Add $5$ to both sides of the equation.

$x = 5$

Step 10

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

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Step 10.1

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

$x^{2} - 5 = 0$

Step 10.2

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

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Step 10.2.1

Add $5$ to both sides of the equation.

$x^{2} = 5$

Step 10.2.2

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

$x = \pm \sqrt{5}$

Step 10.2.3

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

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Step 10.2.3.1

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

$x = \sqrt{5}$

Step 10.2.3.2

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

$x = - \sqrt{5}$

Step 10.2.3.3

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

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

Step 11

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

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

Step 12

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$x = - 5, 5, 2.23606797 \dots, - 2.23606797 \dots$