Find the Roots (Zeros) x^4=8x^2+9

$x^{4} = 8 x^{2} + 9$

Subtract $8 x^{2}$ from both sides of the equation.

$x^{4} - 8 x^{2} = 9$

Subtract $9$ from both sides of the equation.

$x^{4} - 8 x^{2} - 9 = 0$

Factor the left side of the equation.

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

${(x^{2})}^{2} - 8 x^{2} - 9 = 0$

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

$u^{2} - 8 u - 9 = 0$

Factor $u^{2} - 8 u - 9$ 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 $- 9$ and whose sum is $- 8$ .

$- 9, 1$

Write the factored form using these integers.

$(u - 9) (u + 1) = 0$

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

$(x^{2} - 9) (x^{2} + 1) = 0$

Rewrite $9$ as $3^{2}$ .

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

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 $b = 3$ .

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

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

$x + 3 = 0$

$x - 3 = 0$

$x^{2} + 1 = 0$

Set $x + 3$ equal to $0$ and solve for $x$ .

Set $x + 3$ equal to $0$ .

$x + 3 = 0$

Subtract $3$ from both sides of the equation.

$x = - 3$

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

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

$x - 3 = 0$

Add $3$ to both sides of the equation.

$x = 3$

Set $x^{2} + 1$ equal to $0$ and solve for $x$ .

Set $x^{2} + 1$ equal to $0$ .

$x^{2} + 1 = 0$

Solve $x^{2} + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$x^{2} = - 1$

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

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

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

$x = \pm i$

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.

$x = i$

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

$x = - i$

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

$x = i, - i$

The final solution is all the values that make $(x + 3) (x - 3) (x^{2} + 1) = 0$ true.

$x = - 3, 3, i, - i$

Find the Roots (Zeros) x^4=8x^2+9

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