Find the Roots (Zeros) x^4-6x^2+5=0

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

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

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

$u = x^{2}$

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.

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

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

$u - 5 = 0$

$u - 1 = 0$

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

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

$u - 5 = 0$

Add $5$ to both sides of the equation.

$u = 5$

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

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

$u - 1 = 0$

Add $1$ to both sides of the equation.

$u = 1$

The final solution is all the values that make $(u - 5) (u - 1) = 0$ true.

$u = 5, 1$

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = 5$

${(x^{2})}^{1} = 1$

Solve the first equation for $x$ .

$x^{2} = 5$

Solve the equation for $x$ .

Take the square 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.

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

Solve the second equation for $x$ .

${(x^{2})}^{1} = 1$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = 1$

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

$x = \pm \sqrt{1}$

Any root of $1$ is $1$ .

$x = \pm 1$

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.

$x = 1$

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

$x = - 1$

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

$x = 1, - 1$

The solution to $x^{4} - 6 x^{2} + 5 = 0$ is $x = \sqrt{5}, - \sqrt{5}, 1, - 1$ .

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

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

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

Find the Roots (Zeros) x^4-6x^2+5=0

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