Find the Roots (Zeros) 12x^2-64=-x^4

$12 x^{2} - 64 = - x^{4}$

Step 1

Add $x^{4}$ to both sides of the equation.

$12 x^{2} - 64 + x^{4} = 0$

Step 2

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

$u^{2} + 12 u - 64 = 0$

$u = x^{2}$

Step 3

Factor $u^{2} + 12 u - 64$ using the AC method.

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 $- 64$ and whose sum is $12$ .

$- 4, 16$

Write the factored form using these integers.

$(u - 4) (u + 16) = 0$

Step 4

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

$u - 4 = 0$

$u + 16 = 0$

Step 5

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

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

$u - 4 = 0$

Add $4$ to both sides of the equation.

$u = 4$

Step 6

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

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

$u + 16 = 0$

Subtract $16$ from both sides of the equation.

$u = - 16$

Step 7

The final solution is all the values that make $(u - 4) (u + 16) = 0$ true.

$u = 4, - 16$

Step 8

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

$x^{2} = 4$

${(x^{2})}^{1} = - 16$

Step 9

Solve the first equation for $x$ .

$x^{2} = 4$

Step 10

Solve the equation for $x$ .

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

$x = \pm \sqrt{4}$

Simplify $\pm \sqrt{4}$ .

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

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

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

$x = \pm 2$

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

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

$x = 2$

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

$x = - 2$

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

$x = 2, - 2$

Step 11

Solve the second equation for $x$ .

${(x^{2})}^{1} = - 16$

Step 12

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 16$

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

$x = \pm \sqrt{- 16}$

Simplify $\pm \sqrt{- 16}$ .

Rewrite $- 16$ as $- 1 (16)$ .

$x = \pm \sqrt{- 1 (16)}$

Rewrite $\sqrt{- 1 (16)}$ as $\sqrt{- 1} \cdot \sqrt{16}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{16}$

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

$x = \pm i \cdot \sqrt{16}$

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

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

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

$x = \pm i \cdot 4$

Move $4$ to the left of $i$ .

$x = \pm 4 i$

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

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

$x = 4 i$

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

$x = - 4 i$

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

$x = 4 i, - 4 i$

Step 13

The solution to $x^{4} + 12 x^{2} - 64 = 0$ is $x = 2, - 2, 4 i, - 4 i$ .

$x = 2, - 2, 4 i, - 4 i$

Step 14

Find the Roots (Zeros) 12x^2-64=-x^4

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