Precalculus Examples

$f (x) = x^{4} - 6 x^{2} + 8$

Step 1

Set $x^{4} - 6 x^{2} + 8$ equal to $0$ .

$x^{4} - 6 x^{2} + 8 = 0$

Step 2

Solve for $x$ .

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Step 2.1

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

$u^{2} - 6 u + 8 = 0$

$u = x^{2}$

Step 2.2

Factor $u^{2} - 6 u + 8$ using the AC method.

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Step 2.2.1

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 $8$ and whose sum is $- 6$ .

$- 4, - 2$

Step 2.2.2

Write the factored form using these integers.

$(u - 4) (u - 2) = 0$

Step 2.3

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

$u - 4 = 0$

$u - 2 = 0$

Step 2.4

Set $u - 4$ equal to $0$ and solve for $u$ .

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Step 2.4.1

Set $u - 4$ equal to $0$ .

$u - 4 = 0$

Step 2.4.2

Add $4$ to both sides of the equation.

$u = 4$

Step 2.5

Set $u - 2$ equal to $0$ and solve for $u$ .

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Step 2.5.1

Set $u - 2$ equal to $0$ .

$u - 2 = 0$

Step 2.5.2

Add $2$ to both sides of the equation.

$u = 2$

Step 2.6

The final solution is all the values that make $(u - 4) (u - 2) = 0$ true.

$u = 4, 2$

Step 2.7

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

$x^{2} = 4$

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

Step 2.8

Solve the first equation for $x$ .

$x^{2} = 4$

Step 2.9

Solve the equation for $x$ .

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Step 2.9.1

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

$x = \pm \sqrt{4}$

Step 2.9.2

Simplify $\pm \sqrt{4}$ .

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Step 2.9.2.1

Rewrite $4$ as $2^{2}$ .

$x = \pm \sqrt{2^{2}}$

Step 2.9.2.2

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

$x = \pm 2$

Step 2.9.3

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

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Step 2.9.3.1

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

$x = 2$

Step 2.9.3.2

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

$x = - 2$

Step 2.9.3.3

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

$x = 2, - 2$

Step 2.10

Solve the second equation for $x$ .

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

Step 2.11

Solve the equation for $x$ .

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Step 2.11.1

Remove parentheses.

$x^{2} = 2$

Step 2.11.2

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

$x = \pm \sqrt{2}$

Step 2.11.3

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

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Step 2.11.3.1

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

$x = \sqrt{2}$

Step 2.11.3.2

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

$x = - \sqrt{2}$

Step 2.11.3.3

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

$x = \sqrt{2}, - \sqrt{2}$

Step 2.12

The solution to $x^{4} - 6 x^{2} + 8 = 0$ is $x = 2, - 2, \sqrt{2}, - \sqrt{2}$ .

$x = 2, - 2, \sqrt{2}, - \sqrt{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$x = 2, - 2, \sqrt{2}, - \sqrt{2}$

Decimal Form:

$x = 2, - 2, 1.41421356 \dots, - 1.41421356 \dots$

Step 4