Solve by Factoring 4x^4+15x^2-4=0

$4 x^{4} + 15 x^{2} - 4 = 0$

Step 1

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

$4 {(x^{2})}^{2} + 15 x^{2} - 4 = 0$

Step 2

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

$4 u^{2} + 15 u - 4 = 0$

Step 3

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 $a \cdot c = 4 \cdot - 4 = - 16$ and whose sum is $b = 15$ .

Factor $15$ out of $15 u$ .

$4 u^{2} + 15 (u) - 4 = 0$

Rewrite $15$ as $- 1$ plus $16$

$4 u^{2} + (- 1 + 16) u - 4 = 0$

Apply the distributive property.

$4 u^{2} - 1 u + 16 u - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(4 u^{2} - 1 u) + 16 u - 4 = 0$

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

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

Factor the polynomial by factoring out the greatest common factor, $4 u - 1$ .

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

Step 4

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

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

Step 5

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

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

Step 6

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

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

Step 7

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

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

Step 8

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

$2 x + 1 = 0$

$2 x - 1 = 0$

$x^{2} + 4 = 0$

Step 9

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

Set $2 x + 1$ equal to $0$ .

$2 x + 1 = 0$

Solve $2 x + 1 = 0$ for $x$ .

Subtract $1$ from both sides of the equation.

$2 x = - 1$

Divide each term in $2 x = - 1$ by $2$ and simplify.

Divide each term in $2 x = - 1$ by $2$ .

$\frac{2 x}{2} = \frac{- 1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{- 1}{2}$

Divide $x$ by $1$ .

$x = \frac{- 1}{2}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{1}{2}$

Step 10

Set $2 x - 1$ equal to $0$ and solve for $x$ .

Set $2 x - 1$ equal to $0$ .

$2 x - 1 = 0$

Solve $2 x - 1 = 0$ for $x$ .

Add $1$ to both sides of the equation.

$2 x = 1$

Divide each term in $2 x = 1$ by $2$ and simplify.

Divide each term in $2 x = 1$ by $2$ .

$\frac{2 x}{2} = \frac{1}{2}$

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

$\frac{2 x}{2} = \frac{1}{2}$

Divide $x$ by $1$ .

$x = \frac{1}{2}$

Step 11

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 12

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

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

Solve by Factoring 4x^4+15x^2-4=0

Related Questions

Support LCMGCF.com by disabling your adblocker.