Precalculus Examples

$f (x) = x^{4} + 5 x^{2} + 4$

Step 1

Set $x^{4} + 5 x^{2} + 4$ equal to $0$ .

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

Step 2

Solve for $x$ .

Tap for more steps...

Step 2.1

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

$u^{2} + 5 u + 4 = 0$

$u = x^{2}$

Step 2.2

Factor $u^{2} + 5 u + 4$ using the AC method.

Tap for more steps...

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 $4$ and whose sum is $5$ .

$1, 4$

Step 2.2.2

Write the factored form using these integers.

$(u + 1) (u + 4) = 0$

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 $0$ .

$u + 1 = 0$

$u + 4 = 0$

Step 2.4

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

Tap for more steps...

Step 2.4.1

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

$u + 1 = 0$

Step 2.4.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 2.5

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

Tap for more steps...

Step 2.5.1

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

$u + 4 = 0$

Step 2.5.2

Subtract $4$ from both sides of the equation.

$u = - 4$

Step 2.6

The final solution is all the values that make $(u + 1) (u + 4) = 0$ true.

$u = - 1, - 4$

Step 2.7

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

$x^{2} = - 1$

${(x^{2})}^{1} = - 4$

Step 2.8

Solve the first equation for $x$ .

$x^{2} = - 1$

Step 2.9

Solve the equation for $x$ .

Tap for more steps...

Step 2.9.1

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

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

Step 2.9.2

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

$x = \pm i$

Step 2.9.3

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

Tap for more steps...

Step 2.9.3.1

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

$x = i$

Step 2.9.3.2

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

$x = - i$

Step 2.9.3.3

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

$x = i, - i$

Step 2.10

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 4$

Step 2.11

Solve the equation for $x$ .

Tap for more steps...

Step 2.11.1

Remove parentheses.

$x^{2} = - 4$

Step 2.11.2

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

$x = \pm \sqrt{- 4}$

Step 2.11.3

Simplify $\pm \sqrt{- 4}$ .

Tap for more steps...

Step 2.11.3.1

Rewrite $- 4$ as $- 1 (4)$ .

$x = \pm \sqrt{- 1 (4)}$

Step 2.11.3.2

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{4}$

Step 2.11.3.3

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

$x = \pm i \cdot \sqrt{4}$

Step 2.11.3.4

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

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

Step 2.11.3.5

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

$x = \pm i \cdot 2$

Step 2.11.3.6

Move $2$ to the left of $i$ .

$x = \pm 2 i$

Step 2.11.4

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

Tap for more steps...

Step 2.11.4.1

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

$x = 2 i$

Step 2.11.4.2

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

$x = - 2 i$

Step 2.11.4.3

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

$x = 2 i, - 2 i$

Step 2.12

The solution to $x^{4} + 5 x^{2} + 4 = 0$ is $x = i, - i, 2 i, - 2 i$ .

$x = i, - i, 2 i, - 2 i$

Step 3