Solve by Factoring x^4-6x^2+5=0

$x^{4} - 6 x^{2} + 5 = 0$

Step 1

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

${(x^{2})}^{2} - 6 x^{2} + 5 = 0$

Step 2

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

$u^{2} - 6 u + 5 = 0$

Step 3

Factor $u^{2} - 6 u + 5$ using the AC method.

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 $5$ and whose sum is $- 6$ .

$- 5, - 1$

Write the factored form using these integers.

$(u - 5) (u - 1) = 0$

Step 4

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

$(x^{2} - 5) (x^{2} - 1) = 0$

Step 5

Rewrite $1$ as $1^{2}$ .

$(x^{2} - 5) (x^{2} - 1^{2}) = 0$

Step 6

Factor.

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 $b = 1$ .

$(x^{2} - 5) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(x^{2} - 5) (x + 1) (x - 1) = 0$

Step 7

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

$x^{2} - 5 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 8

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 9

Set $x + 1$ equal to $0$ and solve for $x$ .

Set $x + 1$ equal to $0$ .

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 10

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 11

The final solution is all the values that make $(x^{2} - 5) (x + 1) (x - 1) = 0$ true.

$x = \sqrt{5}, - \sqrt{5}, - 1, 1$

Step 12

The result can be shown in multiple forms.

Exact Form:

$x = \sqrt{5}, - \sqrt{5}, - 1, 1$

Decimal Form:

$x = 2.23606797 \dots, - 2.23606797 \dots, - 1, 1$

Solve by Factoring x^4-6x^2+5=0

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