Find All Complex Solutions 0=x^4+3x^2-4

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

Step 1

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

Subtract $x^{4}$ from both sides of the equation.

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

Subtract $3 x^{2}$ from both sides of the equation.

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

Add $4$ to both sides of the equation.

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

Step 2

Factor the left side of the equation.

Factor $- 1$ out of $- x^{4} - 3 x^{2} + 4$ .

Factor $- 1$ out of $- x^{4}$ .

$- (x^{4}) - 3 x^{2} + 4 = 0$

Factor $- 1$ out of $- 3 x^{2}$ .

$- (x^{4}) - (3 x^{2}) + 4 = 0$

Rewrite $4$ as $- 1 (- 4)$ .

$- (x^{4}) - (3 x^{2}) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (x^{4}) - (3 x^{2})$ .

$- (x^{4} + 3 x^{2}) - 1 \cdot - 4 = 0$

Factor $- 1$ out of $- (x^{4} + 3 x^{2}) - 1 (- 4)$ .

$- (x^{4} + 3 x^{2} - 4) = 0$

Rewrite $x^{4}$ as ${(x^{2})}^{2}$ .

$- ({(x^{2})}^{2} + 3 x^{2} - 4) = 0$

Let $u = x^{2}$ . Substitute $u$ for all occurrences of $x^{2}$ .

$- (u^{2} + 3 u - 4) = 0$

Factor $u^{2} + 3 u - 4$ 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 $- 4$ and whose sum is $3$ .

$- 1, 4$

Write the factored form using these integers.

$- ((u - 1) (u + 4)) = 0$

Replace all occurrences of $u$ with $x^{2}$ .

$- ((x^{2} - 1) (x^{2} + 4)) = 0$

Rewrite $1$ as $1^{2}$ .

$- ((x^{2} - 1^{2}) (x^{2} + 4)) = 0$

Factor.

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = x$ and $b = 1$ .

$- ((x + 1) (x - 1) (x^{2} + 4)) = 0$

Remove unnecessary parentheses.

$- (x + 1) (x - 1) (x^{2} + 4) = 0$

Step 3

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

$x + 1 = 0$

$x - 1 = 0$

$x^{2} + 4 = 0$

Step 4

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

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

$x + 1 = 0$

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Set $x - 1$ equal to $0$ and solve for $x$ .

Set $x - 1$ equal to $0$ .

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

Step 6

Set $x^{2} + 4$ equal to $0$ and solve for $x$ .

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

$x^{2} + 4 = 0$

Solve $x^{2} + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$x^{2} = - 4$

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 $- 1 (4)$ .

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

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.

$x = 2 i, - 2 i$

Step 7

The final solution is all the values that make $- (x + 1) (x - 1) (x^{2} + 4) = 0$ true.

$x = - 1, 1, 2 i, - 2 i$

Find All Complex Solutions 0=x^4+3x^2-4

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