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

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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

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

Simplify the left side.

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

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

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

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

Cancel the common factor of $3$ .

Cancel the common factor.

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

Rewrite the expression.

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

Simplify.

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

Simplify the right side.

Simplify ${(2 x)}^{3}$ .

Apply the product rule to $2 x$ .

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

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

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

Step 3

Solve for $x$ .

Subtract $8 x^{3}$ from both sides of the equation.

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

Factor the left side of the equation.

Factor $- 1$ out of $3 - 8 x^{2} - 8 x^{3}$ .

Reorder the expression.

Move $3$ .

$- 8 x^{2} - 8 x^{3} + 3 = 0$

Reorder $- 8 x^{2}$ and $- 8 x^{3}$ .

$- 8 x^{3} - 8 x^{2} + 3 = 0$

Factor $- 1$ out of $- 8 x^{3}$ .

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

Factor $- 1$ out of $- 8 x^{2}$ .

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

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

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

Factor $- 1$ out of $- (8 x^{3}) - (8 x^{2})$ .

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

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

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

Factor.

Factor $8 x^{3} + 8 x^{2} - 3$ using the rational roots test.

If a polynomial function has integer coefficients, then every rational zero will have the form $\frac{p}{q}$ where $p$ is a factor of the constant and $q$ is a factor of the leading coefficient.

$p = \pm 1, \pm 3 q = \pm 1, \pm 8, \pm 2, \pm 4$

Find every combination of $\pm \frac{p}{q}$ . These are the possible roots of the polynomial function.

$\pm 1, \pm 0.125, \pm 0.5, \pm 0.25, \pm 3, \pm 0.375, \pm 1.5, \pm 0.75$

Substitute $0.5$ and simplify the expression. In this case, the expression is equal to $0$ so $0.5$ is a root of the polynomial.

Substitute $0.5$ into the polynomial.

$8 \cdot {0.5}^{3} + 8 \cdot {0.5}^{2} - 3$

Raise $0.5$ to the power of $3$ .

$8 \cdot 0.125 + 8 \cdot {0.5}^{2} - 3$

Multiply $8$ by $0.125$ .

$1 + 8 \cdot {0.5}^{2} - 3$

Raise $0.5$ to the power of $2$ .

$1 + 8 \cdot 0.25 - 3$

Multiply $8$ by $0.25$ .

$1 + 2 - 3$

Add $1$ and $2$ .

$3 - 3$

Subtract $3$ from $3$ .

$0$

Since $0.5$ is a known root, divide the polynomial by $2 x - 1$ to find the quotient polynomial. This polynomial can then be used to find the remaining roots.

$\frac{8 x^{3} + 8 x^{2} - 3}{2 x - 1}$

Divide $8 x^{3} + 8 x^{2} - 3$ by $2 x - 1$ .

Set up the polynomials to be divided. If there is not a term for every exponent, insert one with a value of $0$ .

$2 x$

$1$

$8 x^{3}$

$8 x^{2}$

$0 x$

$3$

Divide the highest order term in the dividend $8 x^{3}$ by the highest order term in divisor $2 x$ .

					$4 x^{2}$
	$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$

Multiply the new quotient term by the divisor.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			+	$8 x^{3}$	-	$4 x^{2}$

The expression needs to be subtracted from the dividend, so change all the signs in $8 x^{3} - 4 x^{2}$

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$

Pull the next terms from the original dividend down into the current dividend.

				$4 x^{2}$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$

Divide the highest order term in the dividend $12 x^{2}$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	+	$6 x$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$6 x$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					+	$12 x^{2}$	-	$6 x$

The expression needs to be subtracted from the dividend, so change all the signs in $12 x^{2} - 6 x$

				$4 x^{2}$	+	$6 x$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$4 x^{2}$	+	$6 x$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$
							+	$6 x$

Pull the next terms from the original dividend down into the current dividend.

				$4 x^{2}$	+	$6 x$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$
							+	$6 x$	-	$3$

Divide the highest order term in the dividend $6 x$ by the highest order term in divisor $2 x$ .

				$4 x^{2}$	+	$6 x$	+	$3$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$
							+	$6 x$	-	$3$

Multiply the new quotient term by the divisor.

				$4 x^{2}$	+	$6 x$	+	$3$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$
							+	$6 x$	-	$3$
							+	$6 x$	-	$3$

The expression needs to be subtracted from the dividend, so change all the signs in $6 x - 3$

				$4 x^{2}$	+	$6 x$	+	$3$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$
							+	$6 x$	-	$3$
							-	$6 x$	+	$3$

After changing the signs, add the last dividend from the multiplied polynomial to find the new dividend.

				$4 x^{2}$	+	$6 x$	+	$3$
$2 x$	-	$1$		$8 x^{3}$	+	$8 x^{2}$	+	$0 x$	-	$3$
			-	$8 x^{3}$	+	$4 x^{2}$
					+	$12 x^{2}$	+	$0 x$
					-	$12 x^{2}$	+	$6 x$
							+	$6 x$	-	$3$
							-	$6 x$	+	$3$
										$0$

Since the remander is $0$ , the final answer is the quotient.

$4 x^{2} + 6 x + 3$

Write $8 x^{3} + 8 x^{2} - 3$ as a set of factors.

$- ((2 x - 1) (4 x^{2} + 6 x + 3)) = 0$

Remove unnecessary parentheses.

$- (2 x - 1) (4 x^{2} + 6 x + 3) = 0$

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

$2 x - 1 = 0$

$4 x^{2} + 6 x + 3 = 0$

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 $1$ .

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

Set $4 x^{2} + 6 x + 3$ equal to $0$ and solve for $x$ .

Set $4 x^{2} + 6 x + 3$ equal to $0$ .

$4 x^{2} + 6 x + 3 = 0$

Solve $4 x^{2} + 6 x + 3 = 0$ for $x$ .

Use the quadratic formula to find the solutions.

$\frac{- b \pm \sqrt{b^{2} - 4 (a c)}}{2 a}$

Substitute the values $a = 4$ , $b = 6$ , and $c = 3$ into the quadratic formula and solve for $x$ .

$\frac{- 6 \pm \sqrt{6^{2} - 4 \cdot (4 \cdot 3)}}{2 \cdot 4}$

Simplify.

Simplify the numerator.

Raise $6$ to the power of $2$ .

$x = \frac{- 6 \pm \sqrt{36 - 4 \cdot 4 \cdot 3}}{2 \cdot 4}$

Multiply $- 4 \cdot 4 \cdot 3$ .

Multiply $- 4$ by $4$ .

$x = \frac{- 6 \pm \sqrt{36 - 16 \cdot 3}}{2 \cdot 4}$

Multiply $- 16$ by $3$ .

$x = \frac{- 6 \pm \sqrt{36 - 48}}{2 \cdot 4}$

Subtract $48$ from $36$ .

$x = \frac{- 6 \pm \sqrt{- 12}}{2 \cdot 4}$

Rewrite $- 12$ as $- 1 (12)$ .

$x = \frac{- 6 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$x = \frac{- 6 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

$x = \frac{- 6 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

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

$x = \frac{- 6 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

Pull terms out from under the radical.

$x = \frac{- 6 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

Move $2$ to the left of $i$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{2 \cdot 4}$

Multiply $2$ by $4$ .

$x = \frac{- 6 \pm 2 i \sqrt{3}}{8}$

Simplify $\frac{- 6 \pm 2 i \sqrt{3}}{8}$ .

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

The final answer is the combination of both solutions.

$x = - \frac{3 - i \sqrt{3}}{4}, - \frac{3 + i \sqrt{3}}{4}$

The final solution is all the values that make $- (2 x - 1) (4 x^{2} + 6 x + 3) = 0$ true.

$x = \frac{1}{2}, - \frac{3 - i \sqrt{3}}{4}, - \frac{3 + i \sqrt{3}}{4}$

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

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