Solve for x cube root of 2x^2+1=-2

$\sqrt[3]{2 x^{2} + 1} = - 2$

Step 1

To remove the radical on the left side of the equation, cube both sides of the equation.

${\sqrt[3]{2 x^{2} + 1}}^{3} = {(- 2)}^{3}$

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

Simplify ${({(2 x^{2} + 1)}^{\frac{1}{3}})}^{3}$ .

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Raise $- 2$ to the power of $3$ .

$2 x^{2} + 1 = - 8$

Step 3

Solve for $x$ .

Move all terms not containing $x$ to the right side of the equation.

Subtract $1$ from both sides of the equation.

$2 x^{2} = - 8 - 1$

Subtract $1$ from $- 8$ .

$2 x^{2} = - 9$

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

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

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

Simplify the left side.

Cancel the common factor of $2$ .

Cancel the common factor.

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

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

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

Simplify the right side.

Move the negative in front of the fraction.

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

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

$x = \pm \sqrt{- \frac{9}{2}}$

Simplify $\pm \sqrt{- \frac{9}{2}}$ .

Rewrite $- \frac{9}{2}$ as $i^{2} \frac{3^{2}}{2}$ .

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

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

Rewrite $9$ as $3^{2}$ .

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

Pull terms out from under the radical.

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

Raise $3$ to the power of $2$ .

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

Rewrite $\sqrt{\frac{9}{2}}$ as $\frac{\sqrt{9}}{\sqrt{2}}$ .

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

Simplify the numerator.

Rewrite $9$ as $3^{2}$ .

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

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

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

Multiply $\frac{3}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Combine and simplify the denominator.

Multiply $\frac{3}{\sqrt{2}}$ by $\frac{\sqrt{2}}{\sqrt{2}}$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Raise $\sqrt{2}$ to the power of $1$ .

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

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Add $1$ and $1$ .

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

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{2}$ as $2^{\frac{1}{2}}$ .

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

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

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

Combine $\frac{1}{2}$ and $2$ .

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

Cancel the common factor of $2$ .

Cancel the common factor.

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

Rewrite the expression.

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

Evaluate the exponent.

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

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

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

Move $3$ to the left of $i$ .

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

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.

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

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

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

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

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

Solve for x cube root of 2x^2+1=-2

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