Algebra Examples

$\sqrt[5]{x^{2} + 2 x} = - 1$

Step 1

Add $1$ to both sides of the equation.

$\sqrt[5]{x^{2} + 2 x} + 1 = 0$

Step 2

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

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Step 2.1

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

$\sqrt[5]{x \cdot x + 2 x} + 1 = 0$

Step 2.2

Factor $x$ out of $2 x$ .

$\sqrt[5]{x \cdot x + x \cdot 2} + 1 = 0$

Step 2.3

Factor $x$ out of $x \cdot x + x \cdot 2$ .

$\sqrt[5]{x (x + 2)} + 1 = 0$

Step 3

Subtract $1$ from both sides of the equation.

$\sqrt[5]{x (x + 2)} = - 1$

Step 4

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $5$ .

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

Step 5

Simplify each side of the equation.

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Step 5.1

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

${({(x (x + 2))}^{\frac{1}{5}})}^{5} = {(- 1)}^{5}$

Step 5.2

Simplify the left side.

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Step 5.2.1

Simplify ${({(x (x + 2))}^{\frac{1}{5}})}^{5}$ .

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Step 5.2.1.1

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

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Step 5.2.1.1.1

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

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

Step 5.2.1.1.2

Cancel the common factor of $5$ .

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Step 5.2.1.1.2.1

Cancel the common factor.

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

Step 5.2.1.1.2.2

Rewrite the expression.

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

Step 5.2.1.2

Apply the distributive property.

${(x \cdot x + x \cdot 2)}^{1} = {(- 1)}^{5}$

Step 5.2.1.3

Simplify the expression.

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Step 5.2.1.3.1

Multiply $x$ by $x$ .

${(x^{2} + x \cdot 2)}^{1} = {(- 1)}^{5}$

Step 5.2.1.3.2

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

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

Step 5.3

Simplify the right side.

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Step 5.3.1

Raise $- 1$ to the power of $5$ .

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

Step 6

Solve for $x$ .

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Step 6.1

Add $1$ to both sides of the equation.

$x^{2} + 2 x + 1 = 0$

Step 6.2

Factor using the perfect square rule.

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Step 6.2.1

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

$x^{2} + 2 x + 1^{2} = 0$

Step 6.2.2

Check that the middle term is two times the product of the numbers being squared in the first term and third term.

$2 x = 2 \cdot x \cdot 1$

Step 6.2.3

Rewrite the polynomial.

$x^{2} + 2 \cdot x \cdot 1 + 1^{2} = 0$

Step 6.2.4

Factor using the perfect square trinomial rule $a^{2} + 2 a b + b^{2} = {(a + b)}^{2}$ , where $a = x$ and $b = 1$ .

${(x + 1)}^{2} = 0$

Step 6.3

Set the $x + 1$ equal to $0$ .

$x + 1 = 0$

Step 6.4

Subtract $1$ from both sides of the equation.

$x = - 1$