Find All Complex Solutions 9x^4-2x^2-7=0

$9 x^{4} - 2 x^{2} - 7 = 0$

Step 1

Factor the left side of the equation.

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

$9 {(x^{2})}^{2} - 2 x^{2} - 7 = 0$

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

$9 u^{2} - 2 u - 7 = 0$

Factor by grouping.

For a polynomial of the form $a x^{2} + b x + c$ , rewrite the middle term as a sum of two terms whose product is $a \cdot c = 9 \cdot - 7 = - 63$ and whose sum is $b = - 2$ .

Factor $- 2$ out of $- 2 u$ .

$9 u^{2} - 2 u - 7 = 0$

Rewrite $- 2$ as $7$ plus $- 9$

$9 u^{2} + (7 - 9) u - 7 = 0$

Apply the distributive property.

$9 u^{2} + 7 u - 9 u - 7 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(9 u^{2} + 7 u) - 9 u - 7 = 0$

Factor out the greatest common factor (GCF) from each group.

$u (9 u + 7) - (9 u + 7) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 u + 7$ .

$(9 u + 7) (u - 1) = 0$

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

$(9 x^{2} + 7) (x^{2} - 1) = 0$

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

$(9 x^{2} + 7) (x^{2} - 1^{2}) = 0$

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 $b = 1$ .

$(9 x^{2} + 7) ((x + 1) (x - 1)) = 0$

Remove unnecessary parentheses.

$(9 x^{2} + 7) (x + 1) (x - 1) = 0$

Step 2

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

$9 x^{2} + 7 = 0$

$x + 1 = 0$

$x - 1 = 0$

Step 3

Set $9 x^{2} + 7$ equal to $0$ and solve for $x$ .

Set $9 x^{2} + 7$ equal to $0$ .

$9 x^{2} + 7 = 0$

Solve $9 x^{2} + 7 = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$9 x^{2} = - 7$

Divide each term in $9 x^{2} = - 7$ by $9$ and simplify.

Divide each term in $9 x^{2} = - 7$ by $9$ .

$\frac{9 x^{2}}{9} = \frac{- 7}{9}$

Simplify the left side.

Cancel the common factor of $9$ .

Cancel the common factor.

$\frac{9 x^{2}}{9} = \frac{- 7}{9}$

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 7}{9}$

Simplify the right side.

Move the negative in front of the fraction.

$x^{2} = - \frac{7}{9}$

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

$x = \pm \sqrt{- \frac{7}{9}}$

Simplify $\pm \sqrt{- \frac{7}{9}}$ .

Rewrite $- \frac{7}{9}$ as ${(\frac{i}{3})}^{2} \cdot 7$ .

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

$x = \pm \sqrt{i^{2} \frac{7}{9}}$

Factor the perfect power $1^{2}$ out of $7$ .

$x = \pm \sqrt{i^{2} \frac{1^{2} \cdot 7}{9}}$

Factor the perfect power $3^{2}$ out of $9$ .

$x = \pm \sqrt{i^{2} \frac{1^{2} \cdot 7}{3^{2} \cdot 1}}$

Rearrange the fraction $\frac{1^{2} \cdot 7}{3^{2} \cdot 1}$ .

$x = \pm \sqrt{i^{2} ({(\frac{1}{3})}^{2} \cdot 7)}$

Rewrite $i^{2} {(\frac{1}{3})}^{2}$ as ${(\frac{i}{3})}^{2}$ .

$x = \pm \sqrt{{(\frac{i}{3})}^{2} \cdot 7}$

Pull terms out from under the radical.

$x = \pm \frac{i}{3} \sqrt{7}$

Combine $\frac{i}{3}$ and $\sqrt{7}$ .

$x = \pm \frac{i \sqrt{7}}{3}$

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.

$x = \frac{i \sqrt{7}}{3}$

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

$x = - \frac{i \sqrt{7}}{3}$

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

$x = \frac{i \sqrt{7}}{3}, - \frac{i \sqrt{7}}{3}$

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

The final solution is all the values that make $(9 x^{2} + 7) (x + 1) (x - 1) = 0$ true.

$x = \frac{i \sqrt{7}}{3}, - \frac{i \sqrt{7}}{3}, - 1, 1$

Find All Complex Solutions 9x^4-2x^2-7=0

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