Precalculus Examples

x^{4} - 3 x^{2} - 28 = 0

$x^{4} - 3 x^{2} - 28 = 0$

Step 1

Substitute

u = x^{2}

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

u^{2} - 3 u - 28 = 0

$u^{2} - 3 u - 28 = 0$

u = x^{2}

$u = x^{2}$

Step 2

Factor

u^{2} - 3 u - 28

$u^{2} - 3 u - 28$ using the AC method.

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Step 2.1

Consider the form

x^{2} + b x + c

$x^{2} + b x + c$ . Find a pair of integers whose product is

c

$c$ and whose sum is

b

$b$ . In this case, whose product is

- 28

$- 28$ and whose sum is

- 3

$- 3$ .

- 7, 4

$- 7, 4$

Step 2.2

Write the factored form using these integers.

(u - 7) (u + 4) = 0

$(u - 7) (u + 4) = 0$

(u - 7) (u + 4) = 0

$(u - 7) (u + 4) = 0$

Step 3

If any individual factor on the left side of the equation is equal to

0

$0$ , the entire expression will be equal to

0

$0$ .

u - 7 = 0

$u - 7 = 0$

u + 4 = 0

$u + 4 = 0$

Step 4

Set

u - 7

$u - 7$ equal to

0

$0$ and solve for

u

$u$ .

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Step 4.1

Set

u - 7

$u - 7$ equal to

0

$0$ .

u - 7 = 0

$u - 7 = 0$

Step 4.2

Add

7

$7$ to both sides of the equation.

u = 7

$u = 7$

u = 7

$u = 7$

Step 5

Set

u + 4

$u + 4$ equal to

0

$0$ and solve for

u

$u$ .

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Step 5.1

Set

u + 4

$u + 4$ equal to

0

$0$ .

u + 4 = 0

$u + 4 = 0$

Step 5.2

Subtract

4

$4$ from both sides of the equation.

u = - 4

$u = - 4$

u = - 4

$u = - 4$

Step 6

The final solution is all the values that make

(u - 7) (u + 4) = 0

$(u - 7) (u + 4) = 0$ true.

u = 7, - 4

$u = 7, - 4$

Step 7

Substitute the real value of

u = x^{2}

$u = x^{2}$ back into the solved equation.

x^{2} = 7

$x^{2} = 7$

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

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

Step 8

Solve the first equation for

x

$x$ .

x^{2} = 7

$x^{2} = 7$

Step 9

Solve the equation for

x

$x$ .

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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{7}

$x = \pm \sqrt{7}$

Step 9.2

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

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Step 9.2.1

First, use the positive value of the

\pm

$\pm$ to find the first solution.

x = \sqrt{7}

$x = \sqrt{7}$

Step 9.2.2

Next, use the negative value of the

\pm

$\pm$ to find the second solution.

x = - \sqrt{7}

$x = - \sqrt{7}$

Step 9.2.3

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

x = \sqrt{7}, - \sqrt{7}

$x = \sqrt{7}, - \sqrt{7}$

x = \sqrt{7}, - \sqrt{7}

$x = \sqrt{7}, - \sqrt{7}$

x = \sqrt{7}, - \sqrt{7}

$x = \sqrt{7}, - \sqrt{7}$

Step 10

Solve the second equation for

x

$x$ .

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

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

Step 11

Solve the equation for

x

$x$ .

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Step 11.1

Remove parentheses.

x^{2} = - 4

$x^{2} = - 4$

Step 11.2

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

x = \pm \sqrt{- 4}

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

Step 11.3

Simplify

\pm \sqrt{- 4}

$\pm \sqrt{- 4}$ .

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Step 11.3.1

Rewrite

$- 4$ as

$- 1 (4)$ .

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

Step 11.3.2

Rewrite

$\sqrt{- 1 (4)}$ as

$\sqrt{- 1} \cdot \sqrt{4}$ .

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

Step 11.3.3

Rewrite

$\sqrt{- 1}$ as

$i$ .

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

Step 11.3.4

Rewrite

$4$ as

$2^{2}$ .

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

Step 11.3.5

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

$x = \pm i \cdot 2$

Step 11.3.6

Move

$2$ to the left of

$i$ .

$x = \pm 2 i$

Step 11.4

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

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Step 11.4.1

First, use the positive value of the

$\pm$ to find the first solution.

$x = 2 i$

Step 11.4.2

Next, use the negative value of the

$\pm$ to find the second solution.

$x = - 2 i$

Step 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 12

The solution to

$x^{4} - 3 x^{2} - 28 = 0$ is

$x = \sqrt{7}, - \sqrt{7}, 2 i, - 2 i$ .

$x = \sqrt{7}, - \sqrt{7}, 2 i, - 2 i$