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

$f (x) = x^{4} + x^{2} - 12$

Step 1

Set $x^{4} + x^{2} - 12$ equal to $0$ .

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

Step 2

Solve for $x$ .

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

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

$u = x^{2}$

Factor $u^{2} + u - 12$ 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 $- 12$ and whose sum is $1$ .

$- 3, 4$

Write the factored form using these integers.

$(u - 3) (u + 4) = 0$

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

$u - 3 = 0$

$u + 4 = 0$

Set $u - 3$ equal to $0$ and solve for $u$ .

Set $u - 3$ equal to $0$ .

$u - 3 = 0$

Add $3$ to both sides of the equation.

$u = 3$

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

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

$u + 4 = 0$

Subtract $4$ from both sides of the equation.

$u = - 4$

The final solution is all the values that make $(u - 3) (u + 4) = 0$ true.

$u = 3, - 4$

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

$x^{2} = 3$

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

Solve the first equation for $x$ .

$x^{2} = 3$

Solve the equation for $x$ .

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

$x = \pm \sqrt{3}$

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.

$x = \sqrt{3}$

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

$x = - \sqrt{3}$

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

$x = \sqrt{3}, - \sqrt{3}$

Solve the second equation for $x$ .

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

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - 4$

Take the square 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 $- 1 (4)$ .

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

Rewrite $\sqrt{- 1 (4)}$ as $\sqrt{- 1} \cdot \sqrt{4}$ .

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

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

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

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

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

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

$x = \pm i \cdot 2$

Move $2$ to the left of $i$ .

$x = \pm 2 i$

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.

$x = 2 i$

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

$x = - 2 i$

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

$x = 2 i, - 2 i$

The solution to $x^{4} + x^{2} - 12 = 0$ is $x = \sqrt{3}, - \sqrt{3}, 2 i, - 2 i$ .

$x = \sqrt{3}, - \sqrt{3}, 2 i, - 2 i$

Step 3

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

Related Questions

Support LCMGCF.com by disabling your adblocker.