Algebra Examples

$\sqrt[4]{x - 1} + 2 = 0$

Step 1

Subtract $2$ from both sides of the equation.

$\sqrt[4]{x - 1} = - 2$

Step 2

To remove the radical on the left side of the equation, raise both sides of the equation to the power of $4$ .

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

Step 3

Simplify each side of the equation.

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Step 3.1

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

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

Step 3.2

Simplify the left side.

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Step 3.2.1

Simplify ${({(x - 1)}^{\frac{1}{4}})}^{4}$ .

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Step 3.2.1.1

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

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $4$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.1.2

Simplify.

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

Step 3.3

Simplify the right side.

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Step 3.3.1

Raise $- 2$ to the power of $4$ .

$x - 1 = 16$

Step 4

Move all terms not containing $x$ to the right side of the equation.

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Step 4.1

Add $1$ to both sides of the equation.

$x = 16 + 1$

Step 4.2

Add $16$ and $1$ .

$x = 17$

Step 5

Exclude the solutions that do not make $\sqrt[4]{x - 1} + 2 = 0$ true.

$No solution$