Solve for x 16x^4+31x^2=2

$16 x^{4} + 31 x^{2} = 2$

Step 1

Subtract $2$ from both sides of the equation.

$16 x^{4} + 31 x^{2} - 2 = 0$

Step 2

Factor the left side of the equation.

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

$16 {(x^{2})}^{2} + 31 x^{2} - 2 = 0$

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

$16 u^{2} + 31 u - 2 = 0$

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 $a \cdot c = 16 \cdot - 2 = - 32$ and whose sum is $b = 31$ .

Factor $31$ out of $31 u$ .

$16 u^{2} + 31 (u) - 2 = 0$

Rewrite $31$ as $- 1$ plus $32$

$16 u^{2} + (- 1 + 32) u - 2 = 0$

Apply the distributive property.

$16 u^{2} - 1 u + 32 u - 2 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(16 u^{2} - 1 u) + 32 u - 2 = 0$

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

$u (16 u - 1) + 2 (16 u - 1) = 0$

Factor the polynomial by factoring out the greatest common factor, $16 u - 1$ .

$(16 u - 1) (u + 2) = 0$

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

$(16 x^{2} - 1) (x^{2} + 2) = 0$

Rewrite $16 x^{2}$ as ${(4 x)}^{2}$ .

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

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

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

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 4 x$ and $b = 1$ .

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

Step 3

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

$4 x + 1 = 0$

$4 x - 1 = 0$

$x^{2} + 2 = 0$

Step 4

Set $4 x + 1$ equal to $0$ and solve for $x$ .

Set $4 x + 1$ equal to $0$ .

$4 x + 1 = 0$

Solve $4 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$4 x = - 1$

Divide each term in $4 x = - 1$ by $4$ and simplify.

Divide each term in $4 x = - 1$ by $4$ .

$\frac{4 x}{4} = \frac{- 1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{- 1}{4}$

Divide $x$ by $1$ .

$x = \frac{- 1}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{4}$

Step 5

Set $4 x - 1$ equal to $0$ and solve for $x$ .

Set $4 x - 1$ equal to $0$ .

$4 x - 1 = 0$

Solve $4 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$4 x = 1$

Divide each term in $4 x = 1$ by $4$ and simplify.

Divide each term in $4 x = 1$ by $4$ .

$\frac{4 x}{4} = \frac{1}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 x}{4} = \frac{1}{4}$

Divide $x$ by $1$ .

$x = \frac{1}{4}$

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

$x = - \frac{1}{4}, \frac{1}{4}, i \sqrt{2}, - i \sqrt{2}$

Solve for x 16x^4+31x^2=2

Related Questions

Support LCMGCF.com by disabling your adblocker.