Precalculus Examples

$f (x) = x^{4} + 6 x^{2} - 27$

Step 1

Set $x^{4} + 6 x^{2} - 27$ equal to $0$ .

$x^{4} + 6 x^{2} - 27 = 0$

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.

$u^{2} + 6 u - 27 = 0$

$u = x^{2}$

Step 2.2

Factor $u^{2} + 6 u - 27$ 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 $- 27$ and whose sum is $6$ .

$- 3, 9$

Step 2.2.2

Write the factored form using these integers.

$(u - 3) (u + 9) = 0$

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 $0$ .

$u - 3 = 0$

$u + 9 = 0$

Step 2.4

Set $u - 3$ equal to $0$ and solve for $u$ .

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Step 2.4.1

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Step 2.4.2

Add $3$ to both sides of the equation.

$u = 3$

Step 2.5

Set $u + 9$ equal to $0$ and solve for $u$ .

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Step 2.5.1

Set $u + 9$ equal to $0$ .

$u + 9 = 0$

Step 2.5.2

Subtract $9$ from both sides of the equation.

$u = - 9$

Step 2.6

The final solution is all the values that make $(u - 3) (u + 9) = 0$ true. The multiplicity of a root is the number of times the root appears.

$u = 3$ (Multiplicity of $1$ )

$u = - 9$ (Multiplicity of $1$ )

Step 2.7

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

$x^{2} = 3$

${(x^{2})}^{1} = - 9$

Step 2.8

Solve the first equation for $x$ .

$x^{2} = 3$

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.

$x = \pm \sqrt{3}$

Step 2.9.2

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

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Step 2.9.2.1

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

$x = \sqrt{3}$

Step 2.9.2.2

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

$x = - \sqrt{3}$

Step 2.9.2.3

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

$x = \sqrt{3}, - \sqrt{3}$

Step 2.9.3

The multiplicity of a root is the number of times the root appears. For example, a factor of ${(x + 5)}^{3}$ would have a root at $x = - 5$ with multiplicity of $3$ .

$x = \sqrt{3}$ (Multiplicity of $1$ )

$x = - \sqrt{3}$ (Multiplicity of $1$ )

$x = \sqrt{3}$ (Multiplicity of $1$ )

$x = - \sqrt{3}$ (Multiplicity of $1$ )

Step 2.10

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 9$

Step 2.11

Solve the equation for $x$ .

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Step 2.11.1

Remove parentheses.

$x^{2} = - 9$

Step 2.11.2

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

$x = \pm \sqrt{- 9}$

Step 2.11.3

Simplify $\pm \sqrt{- 9}$ .

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Step 2.11.3.1

Rewrite $- 9$ as $- 1 (9)$ .

$x = \pm \sqrt{- 1 (9)}$

Step 2.11.3.2

Rewrite $\sqrt{- 1 (9)}$ as $\sqrt{- 1} \cdot \sqrt{9}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{9}$

Step 2.11.3.3

Rewrite $\sqrt{- 1}$ as $i$ .

$x = \pm i \cdot \sqrt{9}$

Step 2.11.3.4

Rewrite $9$ as $3^{2}$ .

$x = \pm i \cdot \sqrt{3^{2}}$

Step 2.11.3.5

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

$x = \pm i \cdot 3$

Step 2.11.3.6

Move $3$ to the left of $i$ .

$x = \pm 3 i$

Step 2.11.4

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

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Step 2.11.4.1

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

$x = 3 i$

Step 2.11.4.2

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

$x = - 3 i$

Step 2.11.4.3

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

$x = 3 i, - 3 i$

Step 2.11.5

The multiplicity of a root is the number of times the root appears. For example, a factor of ${(x + 5)}^{3}$ would have a root at $x = - 5$ with multiplicity of $3$ .

$x = 3 i$ (Multiplicity of $1$ )

$x = - 3 i$ (Multiplicity of $1$ )

$x = 3 i$ (Multiplicity of $1$ )

$x = - 3 i$ (Multiplicity of $1$ )

Step 2.12

The solution to $x^{4} + 6 x^{2} - 27 = 0$ is $x = \sqrt{3}, - \sqrt{3}, 3 i, - 3 i$ .

$x = \sqrt{3}, - \sqrt{3}, 3 i, - 3 i$

Step 3