Precalculus Examples

$x^{4} - x^{2} = 14 x^{2} + 16$

Step 1

Move all the expressions to the left side of the equation.

Tap for more steps...

Step 1.1

Subtract $14 x^{2}$ from both sides of the equation.

$x^{4} - x^{2} - 14 x^{2} = 16$

Step 1.2

Subtract $16$ from both sides of the equation.

$x^{4} - x^{2} - 14 x^{2} - 16 = 0$

Step 2

Subtract $14 x^{2}$ from $- x^{2}$ .

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

Step 3

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 4

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

Tap for more steps...

Step 4.1

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 4.2

Write the factored form using these integers.

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

Step 5

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 6

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

Tap for more steps...

Step 6.1

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

$u - 16 = 0$

Step 6.2

Add $16$ to both sides of the equation.

$u = 16$

Step 7

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

Tap for more steps...

Step 7.1

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

$u + 1 = 0$

Step 7.2

Subtract $1$ from both sides of the equation.

$u = - 1$

Step 8

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

$u = 16, - 1$

Step 9

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

$x^{2} = 16$

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

Step 10

Solve the first equation for $x$ .

$x^{2} = 16$

Step 11

Solve the equation for $x$ .

Tap for more steps...

Step 11.1

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

$x = \pm \sqrt{16}$

Step 11.2

Simplify $\pm \sqrt{16}$ .

Tap for more steps...

Step 11.2.1

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

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

Step 11.2.2

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

$x = \pm 4$

Step 11.3

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

Tap for more steps...

Step 11.3.1

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

$x = 4$

Step 11.3.2

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

$x = - 4$

Step 11.3.3

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

$x = 4, - 4$

Step 12

Solve the second equation for $x$ .

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

Step 13

Solve the equation for $x$ .

Tap for more steps...

Step 13.1

Remove parentheses.

$x^{2} = - 1$

Step 13.2

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

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

Step 13.3

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

$x = \pm i$

Step 13.4

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

Tap for more steps...

Step 13.4.1

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

$x = i$

Step 13.4.2

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

$x = - i$

Step 13.4.3

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

$x = i, - i$

Step 14

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

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