Precalculus Examples

$y^{2} + x + 2 y = 0$

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 $a = 1$ , $b = 2$ , and $c = x$ into the quadratic formula and solve for $y$ .

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

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 $2$ to the power of $2$ .

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

Step 1.3.1.2

Multiply $- 4$ by $1$ .

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

Step 1.3.1.3

Factor $4$ out of $4 - 4 x$ .

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Step 1.3.1.3.1

Factor $4$ out of $4$ .

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

Step 1.3.1.3.2

Factor $4$ out of $- 4 x$ .

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

Step 1.3.1.3.3

Factor $4$ out of $4 (1) + 4 (- x)$ .

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

Step 1.3.1.4

Rewrite $4 (1 - x)$ as $2^{2} (1^{2} - x)$ .

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Step 1.3.1.4.1

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

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

Step 1.3.1.4.2

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

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

Step 1.3.1.5

Pull terms out from under the radical.

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

Step 1.3.1.6

One to any power is one.

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

Step 1.3.2

Multiply $2$ by $1$ .

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

Step 1.3.3

Simplify $\frac{- 2 \pm 2 \sqrt{1 - x}}{2}$ .

$y = - 1 \pm \sqrt{1 - x}$

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 $2$ to the power of $2$ .

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

Step 1.4.1.2

Multiply $- 4$ by $1$ .

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

Step 1.4.1.3

Factor $4$ out of $4 - 4 x$ .

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Step 1.4.1.3.1

Factor $4$ out of $4$ .

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

Step 1.4.1.3.2

Factor $4$ out of $- 4 x$ .

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

Step 1.4.1.3.3

Factor $4$ out of $4 (1) + 4 (- x)$ .

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

Step 1.4.1.4

Rewrite $4 (1 - x)$ as $2^{2} (1^{2} - x)$ .

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Step 1.4.1.4.1

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

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

Step 1.4.1.4.2

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

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

Step 1.4.1.5

Pull terms out from under the radical.

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

Step 1.4.1.6

One to any power is one.

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

Step 1.4.2

Multiply $2$ by $1$ .

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

Step 1.4.3

Simplify $\frac{- 2 \pm 2 \sqrt{1 - x}}{2}$ .

$y = - 1 \pm \sqrt{1 - x}$

Step 1.4.4

Change the $\pm$ to $+$ .

$y = - 1 + \sqrt{1 - x}$

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 $2$ to the power of $2$ .

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

Step 1.5.1.2

Multiply $- 4$ by $1$ .

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

Step 1.5.1.3

Factor $4$ out of $4 - 4 x$ .

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Step 1.5.1.3.1

Factor $4$ out of $4$ .

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

Step 1.5.1.3.2

Factor $4$ out of $- 4 x$ .

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

Step 1.5.1.3.3

Factor $4$ out of $4 (1) + 4 (- x)$ .

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

Step 1.5.1.4

Rewrite $4 (1 - x)$ as $2^{2} (1^{2} - x)$ .

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Step 1.5.1.4.1

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

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

Step 1.5.1.4.2

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

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

Step 1.5.1.5

Pull terms out from under the radical.

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

Step 1.5.1.6

One to any power is one.

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

Step 1.5.2

Multiply $2$ by $1$ .

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

Step 1.5.3

Simplify $\frac{- 2 \pm 2 \sqrt{1 - x}}{2}$ .

$y = - 1 \pm \sqrt{1 - x}$

Step 1.5.4

Change the $\pm$ to $-$ .

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

Step 1.6

The final answer is the combination of both solutions.

$y = - 1 + \sqrt{1 - x}$

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

$y = - 1 + \sqrt{1 - x}$

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

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 $- 1 + \sqrt{1 - x}, - 1 - \sqrt{1 - x}$ .

$y = - 1 + \sqrt{1 - x}$

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

Step 4