Precalculus Examples

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

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 $c = - 8 x + 4$ into the quadratic formula and solve for $y$ .

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

Step 1.3

Simplify.

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Step 1.3.1

Simplify the numerator.

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Step 1.3.1.1

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

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 8 x + 4)}}{2 \cdot 1}$

Step 1.3.1.2

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot (- 8 x + 4)}}{2 \cdot 1}$

Step 1.3.1.3

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{16 - 4 (- 8 x) - 4 \cdot 4}}{2 \cdot 1}$

Step 1.3.1.4

Multiply $- 8$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 32 x - 4 \cdot 4}}{2 \cdot 1}$

Step 1.3.1.5

Multiply $- 4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 + 32 x - 16}}{2 \cdot 1}$

Step 1.3.1.6

Subtract $16$ from $16$ .

$y = \frac{- 4 \pm \sqrt{32 x + 0}}{2 \cdot 1}$

Step 1.3.1.7

Add $32 x$ and $0$ .

$y = \frac{- 4 \pm \sqrt{32 x}}{2 \cdot 1}$

Step 1.3.1.8

Rewrite $32 x$ as ${(2^{2})}^{2} \cdot (2 x)$ .

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Step 1.3.1.8.1

Factor $16$ out of $32$ .

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

Step 1.3.1.8.2

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

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

Step 1.3.1.8.3

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

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

Step 1.3.1.8.4

Add parentheses.

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

Step 1.3.1.9

Pull terms out from under the radical.

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

Step 1.3.1.10

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

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

Step 1.3.2

Multiply $2$ by $1$ .

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

Step 1.3.3

Simplify $\frac{- 4 \pm 4 \sqrt{2 x}}{2}$ .

$y = - 2 \pm 2 \sqrt{2 x}$

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

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

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 8 x + 4)}}{2 \cdot 1}$

Step 1.4.1.2

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot (- 8 x + 4)}}{2 \cdot 1}$

Step 1.4.1.3

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{16 - 4 (- 8 x) - 4 \cdot 4}}{2 \cdot 1}$

Step 1.4.1.4

Multiply $- 8$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 32 x - 4 \cdot 4}}{2 \cdot 1}$

Step 1.4.1.5

Multiply $- 4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 + 32 x - 16}}{2 \cdot 1}$

Step 1.4.1.6

Subtract $16$ from $16$ .

$y = \frac{- 4 \pm \sqrt{32 x + 0}}{2 \cdot 1}$

Step 1.4.1.7

Add $32 x$ and $0$ .

$y = \frac{- 4 \pm \sqrt{32 x}}{2 \cdot 1}$

Step 1.4.1.8

Rewrite $32 x$ as ${(2^{2})}^{2} \cdot (2 x)$ .

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Step 1.4.1.8.1

Factor $16$ out of $32$ .

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

Step 1.4.1.8.2

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

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

Step 1.4.1.8.3

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

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

Step 1.4.1.8.4

Add parentheses.

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

Step 1.4.1.9

Pull terms out from under the radical.

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

Step 1.4.1.10

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

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

Step 1.4.2

Multiply $2$ by $1$ .

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

Step 1.4.3

Simplify $\frac{- 4 \pm 4 \sqrt{2 x}}{2}$ .

$y = - 2 \pm 2 \sqrt{2 x}$

Step 1.4.4

Change the $\pm$ to $+$ .

$y = - 2 + 2 \sqrt{2 x}$

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

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

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot 1 \cdot (- 8 x + 4)}}{2 \cdot 1}$

Step 1.5.1.2

Multiply $- 4$ by $1$ .

$y = \frac{- 4 \pm \sqrt{16 - 4 \cdot (- 8 x + 4)}}{2 \cdot 1}$

Step 1.5.1.3

Apply the distributive property.

$y = \frac{- 4 \pm \sqrt{16 - 4 (- 8 x) - 4 \cdot 4}}{2 \cdot 1}$

Step 1.5.1.4

Multiply $- 8$ by $- 4$ .

$y = \frac{- 4 \pm \sqrt{16 + 32 x - 4 \cdot 4}}{2 \cdot 1}$

Step 1.5.1.5

Multiply $- 4$ by $4$ .

$y = \frac{- 4 \pm \sqrt{16 + 32 x - 16}}{2 \cdot 1}$

Step 1.5.1.6

Subtract $16$ from $16$ .

$y = \frac{- 4 \pm \sqrt{32 x + 0}}{2 \cdot 1}$

Step 1.5.1.7

Add $32 x$ and $0$ .

$y = \frac{- 4 \pm \sqrt{32 x}}{2 \cdot 1}$

Step 1.5.1.8

Rewrite $32 x$ as ${(2^{2})}^{2} \cdot (2 x)$ .

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Step 1.5.1.8.1

Factor $16$ out of $32$ .

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

Step 1.5.1.8.2

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

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

Step 1.5.1.8.3

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

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

Step 1.5.1.8.4

Add parentheses.

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

Step 1.5.1.9

Pull terms out from under the radical.

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

Step 1.5.1.10

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

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

Step 1.5.2

Multiply $2$ by $1$ .

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

Step 1.5.3

Simplify $\frac{- 4 \pm 4 \sqrt{2 x}}{2}$ .

$y = - 2 \pm 2 \sqrt{2 x}$

Step 1.5.4

Change the $\pm$ to $-$ .

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

Step 1.6

The final answer is the combination of both solutions.

$y = - 2 + 2 \sqrt{2 x}$

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

$y = - 2 + 2 \sqrt{2 x}$

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

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 $- 2 + 2 \sqrt{2 x}, - 2 - 2 \sqrt{2 x}$ .

$y = - 2 + 2 \sqrt{2 x}$

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

Step 4