Precalculus Examples

$f (x) = x^{4} + 290 x^{2} + 289$

Step 1

Set $x^{4} + 290 x^{2} + 289$ equal to $0$ .

$x^{4} + 290 x^{2} + 289 = 0$

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.

$u^{2} + 290 u + 289 = 0$

$u = x^{2}$

Step 2.2

Factor $u^{2} + 290 u + 289$ 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 $289$ and whose sum is $290$ .

$1, 289$

Step 2.2.2

Write the factored form using these integers.

$(u + 1) (u + 289) = 0$

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 $0$ .

$u + 1 = 0$

$u + 289 = 0$

Step 2.4

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

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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 $u + 289$ equal to $0$ and solve for $u$ .

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Step 2.5.1

Set $u + 289$ equal to $0$ .

$u + 289 = 0$

Step 2.5.2

Subtract $289$ from both sides of the equation.

$u = - 289$

Step 2.6

The final solution is all the values that make $(u + 1) (u + 289) = 0$ true.

$u = - 1, - 289$

Step 2.7

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

$x^{2} = - 1$

${(x^{2})}^{1} = - 289$

Step 2.8

Solve the first equation for $x$ .

$x^{2} = - 1$

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.

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

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.

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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 $x$ .

${(x^{2})}^{1} = - 289$

Step 2.11

Solve the equation for $x$ .

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Step 2.11.1

Remove parentheses.

$x^{2} = - 289$

Step 2.11.2

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

$x = \pm \sqrt{- 289}$

Step 2.11.3

Simplify $\pm \sqrt{- 289}$ .

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Step 2.11.3.1

Rewrite $- 289$ as $- 1 (289)$ .

$x = \pm \sqrt{- 1 (289)}$

Step 2.11.3.2

Rewrite $\sqrt{- 1 (289)}$ as $\sqrt{- 1} \cdot \sqrt{289}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{289}$

Step 2.11.3.3

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

$x = \pm i \cdot \sqrt{289}$

Step 2.11.3.4

Rewrite $289$ as $17^{2}$ .

$x = \pm i \cdot \sqrt{17^{2}}$

Step 2.11.3.5

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

$x = \pm i \cdot 17$

Step 2.11.3.6

Move $17$ to the left of $i$ .

$x = \pm 17 i$

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.

$x = 17 i$

Step 2.11.4.2

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

$x = - 17 i$

Step 2.11.4.3

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

$x = 17 i, - 17 i$

Step 2.12

The solution to $x^{4} + 290 x^{2} + 289 = 0$ is $x = i, - i, 17 i, - 17 i$ .

$x = i, - i, 17 i, - 17 i$

Step 3