Pre-Algebra Examples

$x = \sqrt{- 1 - 2 x}$

Step 1

Since the radical is on the right side of the equation, switch the sides so it is on the left side of the equation.

$\sqrt{- 1 - 2 x} = x$

Step 2

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

${\sqrt{- 1 - 2 x}}^{2} = x^{2}$

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 $\sqrt{- 1 - 2 x}$ as ${(- 1 - 2 x)}^{\frac{1}{2}}$ .

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

Step 3.2

Simplify the left side.

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Step 3.2.1

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

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Step 3.2.1.1

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

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Step 3.2.1.1.1

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

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

Step 3.2.1.1.2

Cancel the common factor of $2$ .

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Step 3.2.1.1.2.1

Cancel the common factor.

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

Step 3.2.1.1.2.2

Rewrite the expression.

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

Step 3.2.1.2

Simplify.

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

Step 4

Solve for $x$ .

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Step 4.1

Subtract $x^{2}$ from both sides of the equation.

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

Step 4.2

Factor the left side of the equation.

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Step 4.2.1

Factor $- 1$ out of $- 1 - 2 x - x^{2}$ .

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Step 4.2.1.1

Reorder the expression.

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Step 4.2.1.1.1

Move $- 1$ .

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

Step 4.2.1.1.2

Reorder $- 2 x$ and $- x^{2}$ .

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

Step 4.2.1.2

Factor $- 1$ out of $- x^{2}$ .

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

Step 4.2.1.3

Factor $- 1$ out of $- 2 x$ .

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

Step 4.2.1.4

Rewrite $- 1$ as $- 1 (1)$ .

$- (x^{2}) - (2 x) - 1 \cdot 1 = 0$

Step 4.2.1.5

Factor $- 1$ out of $- (x^{2}) - (2 x)$ .

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

Step 4.2.1.6

Factor $- 1$ out of $- (x^{2} + 2 x) - 1 (1)$ .

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

Step 4.2.2

Factor using the perfect square rule.

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Step 4.2.2.1

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

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

Step 4.2.2.2

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 4.2.2.3

Rewrite the polynomial.

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

Step 4.2.2.4

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 4.3

Divide each term in $- {(x + 1)}^{2} = 0$ by $- 1$ and simplify.

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Step 4.3.1

Divide each term in $- {(x + 1)}^{2} = 0$ by $- 1$ .

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

Step 4.3.2

Simplify the left side.

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Step 4.3.2.1

Dividing two negative values results in a positive value.

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

Step 4.3.2.2

Divide ${(x + 1)}^{2}$ by $1$ .

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

Step 4.3.3

Simplify the right side.

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Step 4.3.3.1

Divide $0$ by $- 1$ .

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

Step 4.4

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

$x + 1 = 0$

Step 4.5

Subtract $1$ from both sides of the equation.

$x = - 1$

Step 5

Exclude the solutions that do not make $x = \sqrt{- 1 - 2 x}$ true.

$No solution$