Solve for x fourth root of 3-8x^2=2x

$\sqrt[4]{3 - 8 x^{2}} = 2 x$

Step 1

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $4$ .

${\sqrt[4]{3 - 8 x^{2}}}^{4} = {(2 x)}^{4}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[4]{3 - 8 x^{2}}$ as ${(3 - 8 x^{2})}^{\frac{1}{4}}$ .

${({(3 - 8 x^{2})}^{\frac{1}{4}})}^{4} = {(2 x)}^{4}$

Simplify the left side.

Simplify ${({(3 - 8 x^{2})}^{\frac{1}{4}})}^{4}$ .

Multiply the exponents in ${({(3 - 8 x^{2})}^{\frac{1}{4}})}^{4}$ .

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

${(3 - 8 x^{2})}^{\frac{1}{4} \cdot 4} = {(2 x)}^{4}$

Cancel the common factor of $4$ .

Cancel the common factor.

${(3 - 8 x^{2})}^{\frac{1}{4} \cdot 4} = {(2 x)}^{4}$

Rewrite the expression.

${(3 - 8 x^{2})}^{1} = {(2 x)}^{4}$

Simplify.

$3 - 8 x^{2} = {(2 x)}^{4}$

Simplify the right side.

Simplify ${(2 x)}^{4}$ .

Apply the product rule to $2 x$ .

$3 - 8 x^{2} = 2^{4} x^{4}$

Raise $2$ to the power of $4$ .

$3 - 8 x^{2} = 16 x^{4}$

Step 3

Solve for $x$ .

Subtract $16 x^{4}$ from both sides of the equation.

$3 - 8 x^{2} - 16 x^{4} = 0$

Substitute $u = x^{2}$ into the equation. This will make the quadratic formula easy to use.

$- 16 u^{2} - 8 u + 3 = 0$

$u = x^{2}$

Factor the left side of the equation.

Factor $- 1$ out of $- 16 u^{2} - 8 u + 3$ .

Factor $- 1$ out of $- 16 u^{2}$ .

$- (16 u^{2}) - 8 u + 3 = 0$

Factor $- 1$ out of $- 8 u$ .

$- (16 u^{2}) - (8 u) + 3 = 0$

Rewrite $3$ as $- 1 (- 3)$ .

$- (16 u^{2}) - (8 u) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (16 u^{2}) - (8 u)$ .

$- (16 u^{2} + 8 u) - 1 \cdot - 3 = 0$

Factor $- 1$ out of $- (16 u^{2} + 8 u) - 1 (- 3)$ .

$- (16 u^{2} + 8 u - 3) = 0$

Factor.

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 $a \cdot c = 16 \cdot - 3 = - 48$ and whose sum is $b = 8$ .

Factor $8$ out of $8 u$ .

$- (16 u^{2} + 8 (u) - 3) = 0$

Rewrite $8$ as $- 4$ plus $12$

$- (16 u^{2} + (- 4 + 12) u - 3) = 0$

Apply the distributive property.

$- (16 u^{2} - 4 u + 12 u - 3) = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$- ((16 u^{2} - 4 u) + 12 u - 3) = 0$

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

$- (4 u (4 u - 1) + 3 (4 u - 1)) = 0$

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

$- ((4 u - 1) (4 u + 3)) = 0$

Remove unnecessary parentheses.

$- (4 u - 1) (4 u + 3) = 0$

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

$4 u - 1 = 0$

$4 u + 3 = 0$

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

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

$4 u - 1 = 0$

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

Add $1$ to both sides of the equation.

$4 u = 1$

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

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

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

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

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

Divide $u$ by $1$ .

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

Set $4 u + 3$ equal to $0$ and solve for $u$ .

Set $4 u + 3$ equal to $0$ .

$4 u + 3 = 0$

Solve $4 u + 3 = 0$ for $u$ .

Subtract $3$ from both sides of the equation.

$4 u = - 3$

Divide each term in $4 u = - 3$ by $4$ and simplify.

Divide each term in $4 u = - 3$ by $4$ .

$\frac{4 u}{4} = \frac{- 3}{4}$

Simplify the left side.

Cancel the common factor of $4$ .

Cancel the common factor.

$\frac{4 u}{4} = \frac{- 3}{4}$

Divide $u$ by $1$ .

$u = \frac{- 3}{4}$

Simplify the right side.

Move the negative in front of the fraction.

$u = - \frac{3}{4}$

The final solution is all the values that make $- (4 u - 1) (4 u + 3) = 0$ true.

$u = \frac{1}{4}, - \frac{3}{4}$

Substitute the real value of $u = x^{2}$ back into the solved equation.

$x^{2} = \frac{1}{4}$

${(x^{2})}^{1} = - \frac{3}{4}$

Solve the first equation for $x$ .

$x^{2} = \frac{1}{4}$

Solve the equation for $x$ .

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

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

Simplify $\pm \sqrt{\frac{1}{4}}$ .

Rewrite $\sqrt{\frac{1}{4}}$ as $\frac{\sqrt{1}}{\sqrt{4}}$ .

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

Any root of $1$ is $1$ .

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

Simplify the denominator.

Rewrite $4$ as $2^{2}$ .

$x = \pm \frac{1}{\sqrt{2^{2}}}$

Pull terms out from under the radical, assuming positive real numbers.

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

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.

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

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

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

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

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

Solve the second equation for $x$ .

${(x^{2})}^{1} = - \frac{3}{4}$

Solve the equation for $x$ .

Remove parentheses.

$x^{2} = - \frac{3}{4}$

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

$x = \pm \sqrt{- \frac{3}{4}}$

Simplify $\pm \sqrt{- \frac{3}{4}}$ .

Rewrite $- \frac{3}{4}$ as ${(\frac{i}{2})}^{2} \cdot 3$ .

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

$x = \pm \sqrt{i^{2} \frac{3}{4}}$

Factor the perfect power $1^{2}$ out of $3$ .

$x = \pm \sqrt{i^{2} \frac{1^{2} \cdot 3}{4}}$

Factor the perfect power $2^{2}$ out of $4$ .

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

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

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

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

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

Pull terms out from under the radical.

$x = \pm \frac{i}{2} \sqrt{3}$

Combine $\frac{i}{2}$ and $\sqrt{3}$ .

$x = \pm \frac{i \sqrt{3}}{2}$

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.

$x = \frac{i \sqrt{3}}{2}$

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

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

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

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

The solution to $- 16 x^{4} - 8 x^{2} + 3 = 0$ is $x = \frac{1}{2}, - \frac{1}{2}, \frac{i \sqrt{3}}{2}, - \frac{i \sqrt{3}}{2}$ .

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

Solve for x fourth root of 3-8x^2=2x

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