Finite Math Examples

$x^{4} + 0 x^{3} - 15 x^{2} + 0 x - 16 = 0$

Step 1

Simplify $x^{4} + 0 x^{3} - 15 x^{2} + 0 x - 16$ .

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Step 1.1

Simplify each term.

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Step 1.1.1

Multiply $0$ by $x^{3}$ .

$x^{4} + 0 - 15 x^{2} + 0 x - 16 = 0$

Step 1.1.2

Multiply $0$ by $x$ .

$x^{4} + 0 - 15 x^{2} + 0 - 16 = 0$

Step 1.2

Combine the opposite terms in $x^{4} + 0 - 15 x^{2} + 0 - 16$ .

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Step 1.2.1

Add $x^{4}$ and $0$ .

$x^{4} - 15 x^{2} + 0 - 16 = 0$

Step 1.2.2

Add $x^{4} - 15 x^{2}$ and $0$ .

$x^{4} - 15 x^{2} - 16 = 0$

Step 2

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

$u^{2} - 15 u - 16 = 0$

$u = x^{2}$

Step 3

Factor $u^{2} - 15 u - 16$ using the AC method.

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Step 3.1

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 $- 16$ and whose sum is $- 15$ .

$- 16, 1$

Step 3.2

Write the factored form using these integers.

$(u - 16) (u + 1) = 0$

Step 4

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

$u - 16 = 0$

$u + 1 = 0$

Step 5

Set $u - 16$ equal to $0$ and solve for $u$ .

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Step 5.1

Set $u - 16$ equal to $0$ .

$u - 16 = 0$

Step 5.2

Add $16$ to both sides of the equation.

$u = 16$

Step 6

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

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Step 6.1

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

$u + 1 = 0$

Step 6.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 7

The final solution is all the values that make $(u - 16) (u + 1) = 0$ true.

$u = 16, - 1$

Step 8

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

$x^{2} = 16$

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

Step 9

Solve the first equation for $x$ .

$x^{2} = 16$

Step 10

Solve the equation for $x$ .

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Step 10.1

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

$x = \pm \sqrt{16}$

Step 10.2

Simplify $\pm \sqrt{16}$ .

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Step 10.2.1

Rewrite $16$ as $4^{2}$ .

$x = \pm \sqrt{4^{2}}$

Step 10.2.2

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

$x = \pm 4$

Step 10.3

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

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Step 10.3.1

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

$x = 4$

Step 10.3.2

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

$x = - 4$

Step 10.3.3

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

$x = 4, - 4$

Step 11

Solve the second equation for $x$ .

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

Step 12

Solve the equation for $x$ .

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Step 12.1

Remove parentheses.

$x^{2} = - 1$

Step 12.2

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

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

Step 12.3

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

$x = \pm i$

Step 12.4

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

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Step 12.4.1

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

$x = i$

Step 12.4.2

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

$x = - i$

Step 12.4.3

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

$x = i, - i$

Step 13

The solution to $x^{4} - 15 x^{2} - 16 = 0$ is $x = 4, - 4, i, - i$ .

$x = 4, - 4, i, - i$

Step 14