Pre-Algebra Examples

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

Step 1

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

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

Step 2

Simplify each side of the equation.

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Step 2.1

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

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

Step 2.2

Simplify the left side.

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Step 2.2.1

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

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Step 2.2.1.1

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

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Step 2.2.1.1.1

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

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

Step 2.2.1.1.2

Cancel the common factor of $3$ .

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Step 2.2.1.1.2.1

Cancel the common factor.

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

Step 2.2.1.1.2.2

Rewrite the expression.

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

Step 2.2.1.2

Simplify.

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

Step 2.3

Simplify the right side.

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Step 2.3.1

Rewrite ${\sqrt[3]{4 x}}^{3}$ as $4 x$ .

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Step 2.3.1.1

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

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

Step 2.3.1.2

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

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

Step 2.3.1.3

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

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

Step 2.3.1.4

Cancel the common factor of $3$ .

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Step 2.3.1.4.1

Cancel the common factor.

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

Step 2.3.1.4.2

Rewrite the expression.

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

Step 2.3.1.5

Simplify.

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

Step 3

Solve for $x$ .

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Step 3.1

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

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Step 3.1.1

Subtract $4 x$ from both sides of the equation.

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

Step 3.1.2

Subtract $4 x$ from $3 x$ .

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

Step 3.2

Use the quadratic formula to find the solutions.

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

Step 3.3

Substitute the values $a = 4$ , $b = - 1$ , and $c = - 8$ into the quadratic formula and solve for $x$ .

$\frac{1 \pm \sqrt{{(- 1)}^{2} - 4 \cdot (4 \cdot - 8)}}{2 \cdot 4}$

Step 3.4

Simplify.

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Step 3.4.1

Simplify the numerator.

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Step 3.4.1.1

Raise $- 1$ to the power of $2$ .

$x = \frac{1 \pm \sqrt{1 - 4 \cdot 4 \cdot - 8}}{2 \cdot 4}$

Step 3.4.1.2

Multiply $- 4 \cdot 4 \cdot - 8$ .

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Step 3.4.1.2.1

Multiply $- 4$ by $4$ .

$x = \frac{1 \pm \sqrt{1 - 16 \cdot - 8}}{2 \cdot 4}$

Step 3.4.1.2.2

Multiply $- 16$ by $- 8$ .

$x = \frac{1 \pm \sqrt{1 + 128}}{2 \cdot 4}$

Step 3.4.1.3

Add $1$ and $128$ .

$x = \frac{1 \pm \sqrt{129}}{2 \cdot 4}$

Step 3.4.2

Multiply $2$ by $4$ .

$x = \frac{1 \pm \sqrt{129}}{8}$

Step 3.5

The final answer is the combination of both solutions.

$x = \frac{1 + \sqrt{129}}{8}, \frac{1 - \sqrt{129}}{8}$

Step 4

The result can be shown in multiple forms.

Exact Form:

$x = \frac{1 + \sqrt{129}}{8}, \frac{1 - \sqrt{129}}{8}$

Decimal Form:

$x = 1.54472708 \dots, - 1.29472708 \dots$