Precalculus Examples

$2 x^{2} - 8 y^{3} = 19$ , $4 x^{2} + 16 y^{3} = 34$

Step 1

Simplify the left side.

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Step 1.1

Reorder $2 x^{2}$ and $- 8 y^{3}$ .

$- 8 y^{3} + 2 x^{2} = 19$

$4 x^{2} + 16 y^{3} = 34$

$- 8 y^{3} + 2 x^{2} = 19$

$4 x^{2} + 16 y^{3} = 34$

Step 2

Simplify the left side.

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Step 2.1

Reorder $4 x^{2}$ and $16 y^{3}$ .

$16 y^{3} + 4 x^{2} = 34$

$- 8 y^{3} + 2 x^{2} = 19$

$16 y^{3} + 4 x^{2} = 34$

$- 8 y^{3} + 2 x^{2} = 19$

Step 3

Multiply each equation by the value that makes the coefficients of $x^{2}$ opposite.

$- 2 \cdot (- 8 y^{3} + 2 x^{2}) = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

Step 4

Simplify.

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Step 4.1

Simplify the left side.

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Step 4.1.1

Simplify $- 2 \cdot (- 8 y^{3} + 2 x^{2})$ .

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Step 4.1.1.1

Apply the distributive property.

$- 2 (- 8 y^{3}) - 2 (2 x^{2}) = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

Step 4.1.1.2

Multiply.

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Step 4.1.1.2.1

Multiply $- 8$ by $- 2$ .

$16 y^{3} - 2 (2 x^{2}) = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

Step 4.1.1.2.2

Multiply $2$ by $- 2$ .

$16 y^{3} - 4 x^{2} = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

$16 y^{3} - 4 x^{2} = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

$16 y^{3} - 4 x^{2} = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

$16 y^{3} - 4 x^{2} = (- 2) (19)$

$16 y^{3} + 4 x^{2} = 34$

Step 4.2

Simplify the right side.

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Step 4.2.1

Multiply $- 2$ by $19$ .

$16 y^{3} - 4 x^{2} = - 38$

$16 y^{3} + 4 x^{2} = 34$

$16 y^{3} - 4 x^{2} = - 38$

$16 y^{3} + 4 x^{2} = 34$

$16 y^{3} - 4 x^{2} = - 38$

$16 y^{3} + 4 x^{2} = 34$

Step 5

Add the two equations together to eliminate $x^{2}$ from the system.

	$1$	$6$	$y^{3}$	$-$	$4$	$x^{2}$	$=$	$-$	$3$	$8$
$+$	$1$	$6$	$y^{3}$	$+$	$4$	$x^{2}$	$=$		$3$	$4$
	$3$	$2$	$y^{3}$				$=$		$-$	$4$

Step 6

Divide each term in $32 y^{3} = - 4$ by $32$ and simplify.

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Step 6.1

Divide each term in $32 y^{3} = - 4$ by $32$ .

$\frac{32 y^{3}}{32} = \frac{- 4}{32}$

Step 6.2

Simplify the left side.

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Step 6.2.1

Cancel the common factor of $32$ .

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Step 6.2.1.1

Cancel the common factor.

$\frac{32 y^{3}}{32} = \frac{- 4}{32}$

Step 6.2.1.2

Divide $y^{3}$ by $1$ .

$y^{3} = \frac{- 4}{32}$

Step 6.3

Simplify the right side.

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Step 6.3.1

Cancel the common factor of $- 4$ and $32$ .

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Step 6.3.1.1

Factor $4$ out of $- 4$ .

$y^{3} = \frac{4 (- 1)}{32}$

Step 6.3.1.2

Cancel the common factors.

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Step 6.3.1.2.1

Factor $4$ out of $32$ .

$y^{3} = \frac{4 \cdot - 1}{4 \cdot 8}$

Step 6.3.1.2.2

Cancel the common factor.

$y^{3} = \frac{4 \cdot - 1}{4 \cdot 8}$

Step 6.3.1.2.3

Rewrite the expression.

$y^{3} = \frac{- 1}{8}$

Step 6.3.2

Move the negative in front of the fraction.

$y^{3} = - \frac{1}{8}$

Step 7

Substitute the value found for $y^{3}$ into one of the original equations, then solve for $x^{2}$ .

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Step 7.1

Substitute the value found for $y^{3}$ into one of the original equations to solve for $x^{2}$ .

$16 (- \frac{1}{8}) - 4 x^{2} = - 38$

Step 7.2

Simplify each term.

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Step 7.2.1

Cancel the common factor of $8$ .

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Step 7.2.1.1

Move the leading negative in $- \frac{1}{8}$ into the numerator.

$16 (\frac{- 1}{8}) - 4 x^{2} = - 38$

Step 7.2.1.2

Factor $8$ out of $16$ .

$8 (2) \frac{- 1}{8} - 4 x^{2} = - 38$

Step 7.2.1.3

Cancel the common factor.

$8 \cdot 2 \frac{- 1}{8} - 4 x^{2} = - 38$

Step 7.2.1.4

Rewrite the expression.

$2 \cdot - 1 - 4 x^{2} = - 38$

Step 7.2.2

Multiply $2$ by $- 1$ .

$- 2 - 4 x^{2} = - 38$

Step 7.3

Move all terms not containing $x^{2}$ to the right side of the equation.

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Step 7.3.1

Add $2$ to both sides of the equation.

$- 4 x^{2} = - 38 + 2$

Step 7.3.2

Add $- 38$ and $2$ .

$- 4 x^{2} = - 36$

Step 7.4

Divide each term in $- 4 x^{2} = - 36$ by $- 4$ and simplify.

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Step 7.4.1

Divide each term in $- 4 x^{2} = - 36$ by $- 4$ .

$\frac{- 4 x^{2}}{- 4} = \frac{- 36}{- 4}$

Step 7.4.2

Simplify the left side.

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Step 7.4.2.1

Cancel the common factor of $- 4$ .

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Step 7.4.2.1.1

Cancel the common factor.

$\frac{- 4 x^{2}}{- 4} = \frac{- 36}{- 4}$

Step 7.4.2.1.2

Divide $x^{2}$ by $1$ .

$x^{2} = \frac{- 36}{- 4}$

Step 7.4.3

Simplify the right side.

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Step 7.4.3.1

Divide $- 36$ by $- 4$ .

$x^{2} = 9$

Step 8

This is the final solution to the independent system of equations.

$y^{3} = - \frac{1}{8}$

$x^{2} = 9$

Step 9

Add $\frac{1}{8}$ to both sides of the equation.

$y^{3} + \frac{1}{8} = 0$

$x^{2} = 9$

Step 10

Factor the left side of the equation.

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Step 10.1

Rewrite $1$ as $1^{3}$ .

$y^{3} + \frac{1^{3}}{8} = 0$

$x^{2} = 9$

Step 10.2

Rewrite $8$ as $2^{3}$ .

$y^{3} + \frac{1^{3}}{2^{3}} = 0$

$x^{2} = 9$

Step 10.3

Rewrite $\frac{1^{3}}{2^{3}}$ as ${(\frac{1}{2})}^{3}$ .

$y^{3} + {(\frac{1}{2})}^{3} = 0$

$x^{2} = 9$

Step 10.4

Since both terms are perfect cubes, factor using the sum of cubes formula, $a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$ where $a = y$ and $b = \frac{1}{2}$ .

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

$x^{2} = 9$

Step 10.5

Simplify.

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Step 10.5.1

Combine $\frac{1}{2}$ and $y$ .

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

$x^{2} = 9$

Step 10.5.2

Apply the product rule to $\frac{1}{2}$ .

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

$x^{2} = 9$

Step 10.5.3

One to any power is one.

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

$x^{2} = 9$

Step 10.5.4

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

$(y + \frac{1}{2}) (y^{2} - \frac{y}{2} + \frac{1}{4}) = 0$

$x^{2} = 9$

$(y + \frac{1}{2}) (y^{2} - \frac{y}{2} + \frac{1}{4}) = 0$

$x^{2} = 9$

$(y + \frac{1}{2}) (y^{2} - \frac{y}{2} + \frac{1}{4}) = 0$

$x^{2} = 9$

Step 11

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

$y + \frac{1}{2} = 0$

$y^{2} - \frac{y}{2} + \frac{1}{4} = 0$

$x^{2} = 9$

Step 12

Set $y + \frac{1}{2}$ equal to $0$ and solve for $y$ .

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Step 12.1

Set $y + \frac{1}{2}$ equal to $0$ .

$y + \frac{1}{2} = 0$

$x^{2} = 9$

Step 12.2

Subtract $\frac{1}{2}$ from both sides of the equation.

$y = - \frac{1}{2}$

$x^{2} = 9$

$y = - \frac{1}{2}$

$x^{2} = 9$

Step 13

Set $y^{2} - \frac{y}{2} + \frac{1}{4}$ equal to $0$ and solve for $y$ .

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Step 13.1

Set $y^{2} - \frac{y}{2} + \frac{1}{4}$ equal to $0$ .

$y^{2} - \frac{y}{2} + \frac{1}{4} = 0$

$x^{2} = 9$

Step 13.2

Solve $y^{2} - \frac{y}{2} + \frac{1}{4} = 0$ for $y$ .

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Step 13.2.1

Multiply through by the least common denominator $4$ , then simplify.

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Step 13.2.1.1

Apply the distributive property.

$4 y^{2} + 4 (- \frac{y}{2}) + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2

Simplify.

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Step 13.2.1.2.1

Cancel the common factor of $2$ .

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Step 13.2.1.2.1.1

Move the leading negative in $- \frac{y}{2}$ into the numerator.

$4 y^{2} + 4 (\frac{- y}{2}) + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2.1.2

Factor $2$ out of $4$ .

$4 y^{2} + 2 (2) (\frac{- y}{2}) + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2.1.3

Cancel the common factor.

$4 y^{2} + 2 \cdot (2 (\frac{- y}{2})) + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2.1.4

Rewrite the expression.

$4 y^{2} + 2 (- y) + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

$4 y^{2} + 2 (- y) + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2.2

Multiply $- 1$ by $2$ .

$4 y^{2} - 2 y + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2.3

Cancel the common factor of $4$ .

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Step 13.2.1.2.3.1

Cancel the common factor.

$4 y^{2} - 2 y + 4 (\frac{1}{4}) = 0$

$x^{2} = 9$

Step 13.2.1.2.3.2

Rewrite the expression.

$4 y^{2} - 2 y + 1 = 0$

$x^{2} = 9$

$4 y^{2} - 2 y + 1 = 0$

$x^{2} = 9$

$4 y^{2} - 2 y + 1 = 0$

$x^{2} = 9$

$4 y^{2} - 2 y + 1 = 0$

$x^{2} = 9$

Step 13.2.2

Use the quadratic formula to find the solutions.

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

$x^{2} = 9$

Step 13.2.3

Substitute the values $a = 4$ , $b = - 2$ , and $c = 1$ into the quadratic formula and solve for $y$ .

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

$x^{2} = 9$

Step 13.2.4

Simplify.

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Step 13.2.4.1

Simplify the numerator.

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Step 13.2.4.1.1

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

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

$x^{2} = 9$

Step 13.2.4.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 13.2.4.1.2.1

Multiply $- 4$ by $4$ .

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

$x^{2} = 9$

Step 13.2.4.1.2.2

Multiply $- 16$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.3

Subtract $16$ from $4$ .

$y = \frac{2 \pm \sqrt{- 12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 13.2.4.1.7.1

Factor $4$ out of $12$ .

$y = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.7.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.8

Pull terms out from under the radical.

$y = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.1.9

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

$y = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.4.2

Multiply $2$ by $4$ .

$y = \frac{2 \pm 2 i \sqrt{3}}{8}$

$x^{2} = 9$

Step 13.2.4.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{8}$ .

$y = \frac{1 \pm i \sqrt{3}}{4}$

$x^{2} = 9$

$y = \frac{1 \pm i \sqrt{3}}{4}$

$x^{2} = 9$

Step 13.2.5

Simplify the expression to solve for the $+$ portion of the $\pm$ .

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Step 13.2.5.1

Simplify the numerator.

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Step 13.2.5.1.1

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

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

$x^{2} = 9$

Step 13.2.5.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 13.2.5.1.2.1

Multiply $- 4$ by $4$ .

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

$x^{2} = 9$

Step 13.2.5.1.2.2

Multiply $- 16$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.3

Subtract $16$ from $4$ .

$y = \frac{2 \pm \sqrt{- 12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 13.2.5.1.7.1

Factor $4$ out of $12$ .

$y = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.7.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.8

Pull terms out from under the radical.

$y = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.1.9

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

$y = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.5.2

Multiply $2$ by $4$ .

$y = \frac{2 \pm 2 i \sqrt{3}}{8}$

$x^{2} = 9$

Step 13.2.5.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{8}$ .

$y = \frac{1 \pm i \sqrt{3}}{4}$

$x^{2} = 9$

Step 13.2.5.4

Change the $\pm$ to $+$ .

$y = \frac{1 + i \sqrt{3}}{4}$

$x^{2} = 9$

$y = \frac{1 + i \sqrt{3}}{4}$

$x^{2} = 9$

Step 13.2.6

Simplify the expression to solve for the $-$ portion of the $\pm$ .

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Step 13.2.6.1

Simplify the numerator.

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Step 13.2.6.1.1

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

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

$x^{2} = 9$

Step 13.2.6.1.2

Multiply $- 4 \cdot 4 \cdot 1$ .

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Step 13.2.6.1.2.1

Multiply $- 4$ by $4$ .

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

$x^{2} = 9$

Step 13.2.6.1.2.2

Multiply $- 16$ by $1$ .

$y = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm \sqrt{4 - 16}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.3

Subtract $16$ from $4$ .

$y = \frac{2 \pm \sqrt{- 12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.4

Rewrite $- 12$ as $- 1 (12)$ .

$y = \frac{2 \pm \sqrt{- 1 \cdot 12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.5

Rewrite $\sqrt{- 1 (12)}$ as $\sqrt{- 1} \cdot \sqrt{12}$ .

$y = \frac{2 \pm \sqrt{- 1} \cdot \sqrt{12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.6

Rewrite $\sqrt{- 1}$ as $i$ .

$y = \frac{2 \pm i \cdot \sqrt{12}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.7

Rewrite $12$ as $2^{2} \cdot 3$ .

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Step 13.2.6.1.7.1

Factor $4$ out of $12$ .

$y = \frac{2 \pm i \cdot \sqrt{4 (3)}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.7.2

Rewrite $4$ as $2^{2}$ .

$y = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm i \cdot \sqrt{2^{2} \cdot 3}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.8

Pull terms out from under the radical.

$y = \frac{2 \pm i \cdot (2 \sqrt{3})}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.1.9

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

$y = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

$x^{2} = 9$

$y = \frac{2 \pm 2 i \sqrt{3}}{2 \cdot 4}$

$x^{2} = 9$

Step 13.2.6.2

Multiply $2$ by $4$ .

$y = \frac{2 \pm 2 i \sqrt{3}}{8}$

$x^{2} = 9$

Step 13.2.6.3

Simplify $\frac{2 \pm 2 i \sqrt{3}}{8}$ .

$y = \frac{1 \pm i \sqrt{3}}{4}$

$x^{2} = 9$

Step 13.2.6.4

Change the $\pm$ to $-$ .

$y = \frac{1 - i \sqrt{3}}{4}$

$x^{2} = 9$

$y = \frac{1 - i \sqrt{3}}{4}$

$x^{2} = 9$

Step 13.2.7

The final answer is the combination of both solutions.

$y = \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

$x^{2} = 9$

$y = \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

$x^{2} = 9$

$y = \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

$x^{2} = 9$

Step 14

The final solution is all the values that make $(y + \frac{1}{2}) (y^{2} - \frac{y}{2} + \frac{1}{4}) = 0$ true.

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

$x^{2} = 9$

Step 15

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

$x = \pm \sqrt{9}$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

Step 16

Simplify $\pm \sqrt{9}$ .

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Step 16.1

Rewrite $9$ as $3^{2}$ .

$x = \pm \sqrt{3^{2}}$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

Step 16.2

Pull terms out from under the radical, assuming positive real numbers.

$x = \pm 3$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

$x = \pm 3$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

Step 17

The complete solution is the result of both the positive and negative portions of the solution.

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Step 17.1

First, use the positive value of the $\pm$ to find the first solution.

$x = 3$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

Step 17.2

Next, use the negative value of the $\pm$ to find the second solution.

$x = - 3$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

Step 17.3

The complete solution is the result of both the positive and negative portions of the solution.

$x = 3, - 3$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

$x = 3, - 3$

$y = - \frac{1}{2}, \frac{1 + i \sqrt{3}}{4}, \frac{1 - i \sqrt{3}}{4}$

Step 18

The final result is the combination of all values of $x$ with all values of $y$ .

$(3, - \frac{1}{2}), (3, \frac{1 + i \sqrt{3}}{4}), (3, \frac{1 - i \sqrt{3}}{4}), (- 3, - \frac{1}{2}), (- 3, \frac{1 + i \sqrt{3}}{4}), (- 3, \frac{1 - i \sqrt{3}}{4})$

Step 19