Solve for x cube root of 7x^2=3x

$\sqrt[3]{7 x^{2}} = 3 x$

Step 1

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

${\sqrt[3]{7 x^{2}}}^{3} = {(3 x)}^{3}$

Step 2

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{7 x^{2}}$ as ${(7 x^{2})}^{\frac{1}{3}}$ .

${({(7 x^{2})}^{\frac{1}{3}})}^{3} = {(3 x)}^{3}$

Simplify the left side.

Simplify ${({(7 x^{2})}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(7 x^{2})}^{\frac{1}{3}})}^{3}$ .

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

${(7 x^{2})}^{\frac{1}{3} \cdot 3} = {(3 x)}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(7 x^{2})}^{\frac{1}{3} \cdot 3} = {(3 x)}^{3}$

Rewrite the expression.

${(7 x^{2})}^{1} = {(3 x)}^{3}$

Simplify.

$7 x^{2} = {(3 x)}^{3}$

Simplify the right side.

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

Apply the product rule to $3 x$ .

$7 x^{2} = 3^{3} x^{3}$

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

$7 x^{2} = 27 x^{3}$

Step 3

Solve for $x$ .

Subtract $27 x^{3}$ from both sides of the equation.

$7 x^{2} - 27 x^{3} = 0$

Factor $x^{2}$ out of $7 x^{2} - 27 x^{3}$ .

Factor $x^{2}$ out of $7 x^{2}$ .

$x^{2} \cdot 7 - 27 x^{3} = 0$

Factor $x^{2}$ out of $- 27 x^{3}$ .

$x^{2} \cdot 7 + x^{2} (- 27 x) = 0$

Factor $x^{2}$ out of $x^{2} \cdot 7 + x^{2} (- 27 x)$ .

$x^{2} (7 - 27 x) = 0$

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

$x^{2} = 0$

$7 - 27 x = 0$

Set $x^{2}$ equal to $0$ and solve for $x$ .

Set $x^{2}$ equal to $0$ .

$x^{2} = 0$

Solve $x^{2} = 0$ for $x$ .

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

$x = \pm \sqrt{0}$

Simplify $\pm \sqrt{0}$ .

Rewrite $0$ as $0^{2}$ .

$x = \pm \sqrt{0^{2}}$

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

$x = \pm 0$

Plus or minus $0$ is $0$ .

$x = 0$

Set $7 - 27 x$ equal to $0$ and solve for $x$ .

Set $7 - 27 x$ equal to $0$ .

$7 - 27 x = 0$

Solve $7 - 27 x = 0$ for $x$ .

Subtract $7$ from both sides of the equation.

$- 27 x = - 7$

Divide each term in $- 27 x = - 7$ by $- 27$ and simplify.

Divide each term in $- 27 x = - 7$ by $- 27$ .

$\frac{- 27 x}{- 27} = \frac{- 7}{- 27}$

Simplify the left side.

Cancel the common factor of $- 27$ .

Cancel the common factor.

$\frac{- 27 x}{- 27} = \frac{- 7}{- 27}$

Divide $x$ by $1$ .

$x = \frac{- 7}{- 27}$

Simplify the right side.

Dividing two negative values results in a positive value.

$x = \frac{7}{27}$

The final solution is all the values that make $x^{2} (7 - 27 x) = 0$ true.

$x = 0, \frac{7}{27}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = 0, \frac{7}{27}$

Decimal Form:

$x = 0, 0.\overline{259}$

Solve for x cube root of 7x^2=3x

Related Questions

Support LCMGCF.com by disabling your adblocker.