Solve for x 9x^4+17x^2=2

$9 x^{4} + 17 x^{2} = 2$

Step 1

Subtract $2$ from both sides of the equation.

$9 x^{4} + 17 x^{2} - 2 = 0$

Step 2

Factor the left side of the equation.

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

$9 {(x^{2})}^{2} + 17 x^{2} - 2 = 0$

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

$9 u^{2} + 17 u - 2 = 0$

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 $a \cdot c = 9 \cdot - 2 = - 18$ and whose sum is $b = 17$ .

Factor $17$ out of $17 u$ .

$9 u^{2} + 17 (u) - 2 = 0$

Rewrite $17$ as $- 1$ plus $18$

$9 u^{2} + (- 1 + 18) u - 2 = 0$

Apply the distributive property.

$9 u^{2} - 1 u + 18 u - 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(9 u^{2} - 1 u) + 18 u - 2 = 0$

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

$u (9 u - 1) + 2 (9 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $9 u - 1$ .

$(9 u - 1) (u + 2) = 0$

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

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

Rewrite $9 x^{2}$ as ${(3 x)}^{2}$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 3 x$ and $b = 1$ .

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

Step 3

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

$3 x + 1 = 0$

$3 x - 1 = 0$

$x^{2} + 2 = 0$

Step 4

Set $3 x + 1$ equal to $0$ and solve for $x$ .

Set $3 x + 1$ equal to $0$ .

$3 x + 1 = 0$

Solve $3 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$3 x = - 1$

Divide each term in $3 x = - 1$ by $3$ and simplify.

Divide each term in $3 x = - 1$ by $3$ .

$\frac{3 x}{3} = \frac{- 1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{- 1}{3}$

Divide $x$ by $1$ .

$x = \frac{- 1}{3}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{3}$

Step 5

Set $3 x - 1$ equal to $0$ and solve for $x$ .

Set $3 x - 1$ equal to $0$ .

$3 x - 1 = 0$

Solve $3 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$3 x = 1$

Divide each term in $3 x = 1$ by $3$ and simplify.

Divide each term in $3 x = 1$ by $3$ .

$\frac{3 x}{3} = \frac{1}{3}$

Simplify the left side.

Cancel the common factor of $3$ .

Cancel the common factor.

$\frac{3 x}{3} = \frac{1}{3}$

Divide $x$ by $1$ .

$x = \frac{1}{3}$

Step 6

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

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

$x^{2} + 2 = 0$

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

Subtract $2$ from both sides of the equation.

$x^{2} = - 2$

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

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

Simplify $\pm \sqrt{- 2}$ .

Rewrite $- 2$ as $- 1 (2)$ .

$x = \pm \sqrt{- 1 (2)}$

Rewrite $\sqrt{- 1 (2)}$ as $\sqrt{- 1} \cdot \sqrt{2}$ .

$x = \pm \sqrt{- 1} \cdot \sqrt{2}$

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

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

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.

$x = i \sqrt{2}$

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

$x = - i \sqrt{2}$

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

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

Step 7

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

$x = - \frac{1}{3}, \frac{1}{3}, i \sqrt{2}, - i \sqrt{2}$

Solve for x 9x^4+17x^2=2

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