Solve for x cube root of 5x^2+8x-4- cube root of 9x=0

$\sqrt[3]{5 x^{2} + 8 x - 4} - \sqrt[3]{9 x} = 0$

Step 1

Add $\sqrt[3]{9 x}$ to both sides of the equation.

$\sqrt[3]{5 x^{2} + 8 x - 4} = \sqrt[3]{9 x}$

Step 2

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

${\sqrt[3]{5 x^{2} + 8 x - 4}}^{3} = {\sqrt[3]{9 x}}^{3}$

Step 3

Simplify each side of the equation.

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{5 x^{2} + 8 x - 4}$ as ${(5 x^{2} + 8 x - 4)}^{\frac{1}{3}}$ .

${({(5 x^{2} + 8 x - 4)}^{\frac{1}{3}})}^{3} = {\sqrt[3]{9 x}}^{3}$

Simplify the left side.

Simplify ${({(5 x^{2} + 8 x - 4)}^{\frac{1}{3}})}^{3}$ .

Multiply the exponents in ${({(5 x^{2} + 8 x - 4)}^{\frac{1}{3}})}^{3}$ .

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

${(5 x^{2} + 8 x - 4)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{9 x}}^{3}$

Cancel the common factor of $3$ .

Cancel the common factor.

${(5 x^{2} + 8 x - 4)}^{\frac{1}{3} \cdot 3} = {\sqrt[3]{9 x}}^{3}$

Rewrite the expression.

${(5 x^{2} + 8 x - 4)}^{1} = {\sqrt[3]{9 x}}^{3}$

Simplify.

$5 x^{2} + 8 x - 4 = {\sqrt[3]{9 x}}^{3}$

Simplify the right side.

Rewrite ${\sqrt[3]{9 x}}^{3}$ as $9 x$ .

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt[3]{9 x}$ as ${(9 x)}^{\frac{1}{3}}$ .

$5 x^{2} + 8 x - 4 = {({(9 x)}^{\frac{1}{3}})}^{3}$

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

$5 x^{2} + 8 x - 4 = {(9 x)}^{\frac{1}{3} \cdot 3}$

Combine $\frac{1}{3}$ and $3$ .

$5 x^{2} + 8 x - 4 = {(9 x)}^{\frac{3}{3}}$

Cancel the common factor of $3$ .

Cancel the common factor.

$5 x^{2} + 8 x - 4 = {(9 x)}^{\frac{3}{3}}$

Rewrite the expression.

$5 x^{2} + 8 x - 4 = {(9 x)}^{1}$

Simplify.

$5 x^{2} + 8 x - 4 = 9 x$

Step 4

Solve for $x$ .

Move all terms containing $x$ to the left side of the equation.

Subtract $9 x$ from both sides of the equation.

$5 x^{2} + 8 x - 4 - 9 x = 0$

Subtract $9 x$ from $8 x$ .

$5 x^{2} - x - 4 = 0$

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 $a \cdot c = 5 \cdot - 4 = - 20$ and whose sum is $b = - 1$ .

Factor $- 1$ out of $- x$ .

$5 x^{2} - (x) - 4 = 0$

Rewrite $- 1$ as $4$ plus $- 5$

$5 x^{2} + (4 - 5) x - 4 = 0$

Apply the distributive property.

$5 x^{2} + 4 x - 5 x - 4 = 0$

Factor out the greatest common factor from each group.

Group the first two terms and the last two terms.

$(5 x^{2} + 4 x) - 5 x - 4 = 0$

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

$x (5 x + 4) - (5 x + 4) = 0$

Factor the polynomial by factoring out the greatest common factor, $5 x + 4$ .

$(5 x + 4) (x - 1) = 0$

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

$5 x + 4 = 0$

$x - 1 = 0$

Set $5 x + 4$ equal to $0$ and solve for $x$ .

Set $5 x + 4$ equal to $0$ .

$5 x + 4 = 0$

Solve $5 x + 4 = 0$ for $x$ .

Subtract $4$ from both sides of the equation.

$5 x = - 4$

Divide each term in $5 x = - 4$ by $5$ and simplify.

Divide each term in $5 x = - 4$ by $5$ .

$\frac{5 x}{5} = \frac{- 4}{5}$

Simplify the left side.

Cancel the common factor of $5$ .

Cancel the common factor.

$\frac{5 x}{5} = \frac{- 4}{5}$

Divide $x$ by $1$ .

$x = \frac{- 4}{5}$

Simplify the right side.

Move the negative in front of the fraction.

$x = - \frac{4}{5}$

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

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

$x - 1 = 0$

Add $1$ to both sides of the equation.

$x = 1$

The final solution is all the values that make $(5 x + 4) (x - 1) = 0$ true.

$x = - \frac{4}{5}, 1$

Step 5

The result can be shown in multiple forms.

Exact Form:

$x = - \frac{4}{5}, 1$

Decimal Form:

$x = - 0.8, 1$

Solve for x cube root of 5x^2+8x-4- cube root of 9x=0

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