Precalculus Examples

$x^{4} - 13 x^{2} + 40 = 0$

Step 1

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

$u^{2} - 13 u + 40 = 0$

$u = x^{2}$

Step 2

Factor $u^{2} - 13 u + 40$ using the AC method.

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Step 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 $40$ and whose sum is $- 13$ .

$- 8, - 5$

Step 2.2

Write the factored form using these integers.

$(u - 8) (u - 5) = 0$

Step 3

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

$u - 8 = 0$

$u - 5 = 0$

Step 4

Set $u - 8$ equal to $0$ and solve for $u$ .

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Step 4.1

Set $u - 8$ equal to $0$ .

$u - 8 = 0$

Step 4.2

Add $8$ to both sides of the equation.

$u = 8$

Step 5

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

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Step 5.1

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

$u - 5 = 0$

Step 5.2

Add $5$ to both sides of the equation.

$u = 5$

Step 6

The final solution is all the values that make $(u - 8) (u - 5) = 0$ true.

$u = 8, 5$

Step 7

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

$x^{2} = 8$

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

Step 8

Solve the first equation for $x$ .

$x^{2} = 8$

Step 9

Solve the equation for $x$ .

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Step 9.1

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

$x = \pm \sqrt{8}$

Step 9.2

Simplify $\pm \sqrt{8}$ .

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Step 9.2.1

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 9.2.1.1

Factor $4$ out of $8$ .

$x = \pm \sqrt{4 (2)}$

Step 9.2.1.2

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

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

Step 9.2.2

Pull terms out from under the radical.

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

Step 9.3

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

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Step 9.3.1

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

$x = 2 \sqrt{2}$

Step 9.3.2

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

$x = - 2 \sqrt{2}$

Step 9.3.3

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

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

Step 10

Solve the second equation for $x$ .

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

Step 11

Solve the equation for $x$ .

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Step 11.1

Remove parentheses.

$x^{2} = 5$

Step 11.2

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

$x = \pm \sqrt{5}$

Step 11.3

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

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Step 11.3.1

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

$x = \sqrt{5}$

Step 11.3.2

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

$x = - \sqrt{5}$

Step 11.3.3

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

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

Step 12

The solution to $x^{4} - 13 x^{2} + 40 = 0$ is $x = 2 \sqrt{2}, - 2 \sqrt{2}, \sqrt{5}, - \sqrt{5}$ .

$x = 2 \sqrt{2}, - 2 \sqrt{2}, \sqrt{5}, - \sqrt{5}$

Step 13

The result can be shown in multiple forms.

Exact Form:

$x = 2 \sqrt{2}, - 2 \sqrt{2}, \sqrt{5}, - \sqrt{5}$

Decimal Form:

$x = 2.82842712 \dots, - 2.82842712 \dots, 2.23606797 \dots, - 2.23606797 \dots$