Precalculus Examples

$y^{2} - 4 y + 12 x - 8 = 0$

Step 1

Solve for $y$ .

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Step 1.1

Use the quadratic formula to find the solutions.

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

Step 1.2

Substitute the values $a = 1$ , $b = - 4$ , and $c = 12 x - 8$ into the quadratic formula and solve for $y$ .

$\frac{4 \pm \sqrt{{(- 4)}^{2} - 4 \cdot (1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.3

Simplify.

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Step 1.3.1

Simplify the numerator.

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Step 1.3.1.1

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (12 x - 8)$ .

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Step 1.3.1.1.1

Factor $- 4$ out of ${(- 4)}^{2}$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 1 \cdot (12 x - 8)}}{2 \cdot 1}$

Step 1.3.1.1.2

Factor $- 4$ out of $- 4 \cdot 1 \cdot (12 x - 8)$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.3.1.1.3

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (12 x - 8))$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.3.1.2

Multiply $12 x - 8$ by $1$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 12 x - 8)}}{2 \cdot 1}$

Step 1.3.1.3

Subtract $8$ from $- 4$ .

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

Step 1.3.1.4

Factor $12$ out of $12 x - 12$ .

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Step 1.3.1.4.1

Factor $12$ out of $12 x$ .

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

Step 1.3.1.4.2

Factor $12$ out of $- 12$ .

$y = \frac{4 \pm \sqrt{- 4 (12 (x) + 12 (- 1))}}{2 \cdot 1}$

Step 1.3.1.4.3

Factor $12$ out of $12 (x) + 12 (- 1)$ .

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

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

Step 1.3.1.5

Multiply $- 4$ by $12$ .

$y = \frac{4 \pm \sqrt{- 48 (x - 1)}}{2 \cdot 1}$

Step 1.3.1.6

Rewrite $- 48 (x - 1)$ as ${(2^{2})}^{2} \cdot (- 3 (x - 1))$ .

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Step 1.3.1.6.1

Factor $16$ out of $- 48$ .

$y = \frac{4 \pm \sqrt{16 (- 3) (x - 1)}}{2 \cdot 1}$

Step 1.3.1.6.2

Rewrite $16$ as $4^{2}$ .

$y = \frac{4 \pm \sqrt{4^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.3.1.6.3

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

$y = \frac{4 \pm \sqrt{{(2^{2})}^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.3.1.6.4

Add parentheses.

$y = \frac{4 \pm \sqrt{{(2^{2})}^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.3.1.7

Pull terms out from under the radical.

$y = \frac{4 \pm 2^{2} \sqrt{- 3 (x - 1)}}{2 \cdot 1}$

Step 1.3.1.8

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

$y = \frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2 \cdot 1}$

Step 1.3.2

Multiply $2$ by $1$ .

$y = \frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2}$

Step 1.3.3

Simplify $\frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2}$ .

$y = 2 \pm 2 \sqrt{- 3 (x - 1)}$

Step 1.4

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

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Step 1.4.1

Simplify the numerator.

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Step 1.4.1.1

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (12 x - 8)$ .

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Step 1.4.1.1.1

Factor $- 4$ out of ${(- 4)}^{2}$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 1 \cdot (12 x - 8)}}{2 \cdot 1}$

Step 1.4.1.1.2

Factor $- 4$ out of $- 4 \cdot 1 \cdot (12 x - 8)$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.4.1.1.3

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (12 x - 8))$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.4.1.2

Multiply $12 x - 8$ by $1$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 12 x - 8)}}{2 \cdot 1}$

Step 1.4.1.3

Subtract $8$ from $- 4$ .

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

Step 1.4.1.4

Factor $12$ out of $12 x - 12$ .

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Step 1.4.1.4.1

Factor $12$ out of $12 x$ .

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

Step 1.4.1.4.2

Factor $12$ out of $- 12$ .

$y = \frac{4 \pm \sqrt{- 4 (12 (x) + 12 (- 1))}}{2 \cdot 1}$

Step 1.4.1.4.3

Factor $12$ out of $12 (x) + 12 (- 1)$ .

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

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

Step 1.4.1.5

Multiply $- 4$ by $12$ .

$y = \frac{4 \pm \sqrt{- 48 (x - 1)}}{2 \cdot 1}$

Step 1.4.1.6

Rewrite $- 48 (x - 1)$ as ${(2^{2})}^{2} \cdot (- 3 (x - 1))$ .

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Step 1.4.1.6.1

Factor $16$ out of $- 48$ .

$y = \frac{4 \pm \sqrt{16 (- 3) (x - 1)}}{2 \cdot 1}$

Step 1.4.1.6.2

Rewrite $16$ as $4^{2}$ .

$y = \frac{4 \pm \sqrt{4^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.4.1.6.3

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

$y = \frac{4 \pm \sqrt{{(2^{2})}^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.4.1.6.4

Add parentheses.

$y = \frac{4 \pm \sqrt{{(2^{2})}^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.4.1.7

Pull terms out from under the radical.

$y = \frac{4 \pm 2^{2} \sqrt{- 3 (x - 1)}}{2 \cdot 1}$

Step 1.4.1.8

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

$y = \frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2 \cdot 1}$

Step 1.4.2

Multiply $2$ by $1$ .

$y = \frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2}$

Step 1.4.3

Simplify $\frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2}$ .

$y = 2 \pm 2 \sqrt{- 3 (x - 1)}$

Step 1.4.4

Change the $\pm$ to $+$ .

$y = 2 + 2 \sqrt{- 3 (x - 1)}$

Step 1.5

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

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Step 1.5.1

Simplify the numerator.

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Step 1.5.1.1

Factor $- 4$ out of ${(- 4)}^{2} - 4 \cdot 1 \cdot (12 x - 8)$ .

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Step 1.5.1.1.1

Factor $- 4$ out of ${(- 4)}^{2}$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 \cdot 1 \cdot (12 x - 8)}}{2 \cdot 1}$

Step 1.5.1.1.2

Factor $- 4$ out of $- 4 \cdot 1 \cdot (12 x - 8)$ .

$y = \frac{4 \pm \sqrt{- 4 \cdot - 4 - 4 (1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.5.1.1.3

Factor $- 4$ out of $- 4 \cdot - 4 - 4 (1 \cdot (12 x - 8))$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 1 \cdot (12 x - 8))}}{2 \cdot 1}$

Step 1.5.1.2

Multiply $12 x - 8$ by $1$ .

$y = \frac{4 \pm \sqrt{- 4 (- 4 + 12 x - 8)}}{2 \cdot 1}$

Step 1.5.1.3

Subtract $8$ from $- 4$ .

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

Step 1.5.1.4

Factor $12$ out of $12 x - 12$ .

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Step 1.5.1.4.1

Factor $12$ out of $12 x$ .

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

Step 1.5.1.4.2

Factor $12$ out of $- 12$ .

$y = \frac{4 \pm \sqrt{- 4 (12 (x) + 12 (- 1))}}{2 \cdot 1}$

Step 1.5.1.4.3

Factor $12$ out of $12 (x) + 12 (- 1)$ .

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

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

Step 1.5.1.5

Multiply $- 4$ by $12$ .

$y = \frac{4 \pm \sqrt{- 48 (x - 1)}}{2 \cdot 1}$

Step 1.5.1.6

Rewrite $- 48 (x - 1)$ as ${(2^{2})}^{2} \cdot (- 3 (x - 1))$ .

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Step 1.5.1.6.1

Factor $16$ out of $- 48$ .

$y = \frac{4 \pm \sqrt{16 (- 3) (x - 1)}}{2 \cdot 1}$

Step 1.5.1.6.2

Rewrite $16$ as $4^{2}$ .

$y = \frac{4 \pm \sqrt{4^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.5.1.6.3

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

$y = \frac{4 \pm \sqrt{{(2^{2})}^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.5.1.6.4

Add parentheses.

$y = \frac{4 \pm \sqrt{{(2^{2})}^{2} \cdot (- 3 (x - 1))}}{2 \cdot 1}$

Step 1.5.1.7

Pull terms out from under the radical.

$y = \frac{4 \pm 2^{2} \sqrt{- 3 (x - 1)}}{2 \cdot 1}$

Step 1.5.1.8

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

$y = \frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2 \cdot 1}$

Step 1.5.2

Multiply $2$ by $1$ .

$y = \frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2}$

Step 1.5.3

Simplify $\frac{4 \pm 4 \sqrt{- 3 (x - 1)}}{2}$ .

$y = 2 \pm 2 \sqrt{- 3 (x - 1)}$

Step 1.5.4

Change the $\pm$ to $-$ .

$y = 2 - 2 \sqrt{- 3 (x - 1)}$

Step 1.6

The final answer is the combination of both solutions.

$y = 2 + 2 \sqrt{- 3 (x - 1)}$

$y = 2 - 2 \sqrt{- 3 (x - 1)}$

$y = 2 + 2 \sqrt{- 3 (x - 1)}$

$y = 2 - 2 \sqrt{- 3 (x - 1)}$

Step 2

To write a polynomial in standard form, simplify and then arrange the terms in descending order.

$y = a x^{2} + b x + c$

Step 3

The standard form is $2 + 2 \sqrt{- 3 (x - 1)}, 2 - 2 \sqrt{- 3 (x - 1)}$ .

$y = 2 + 2 \sqrt{- 3 (x - 1)}$

$y = 2 - 2 \sqrt{- 3 (x - 1)}$

Step 4