Calculus Examples

x = 2 y - y^{2}

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

Step 1

Rewrite the equation as

2 y - y^{2} = x

$2 y - y^{2} = x$ .

2 y - y^{2} = x

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

Step 2

Subtract

x

$x$ from both sides of the equation.

2 y - y^{2} - x = 0

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

Step 3

Use the quadratic formula to find the solutions.

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

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

Step 4

Substitute the values

a = - 1

$a = - 1$ ,

b = 2

$b = 2$ , and

c = - x

$c = - x$ into the quadratic formula and solve for

y

$y$ .

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

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

Step 5

Simplify.

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Step 5.1

Simplify the numerator.

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Step 5.1.1

Raise

2

$2$ to the power of

2

$2$ .

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

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

Step 5.1.2

Multiply

- 4 \cdot - 1 \cdot - 1

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

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Step 5.1.2.1

Multiply

- 4

$- 4$ by

- 1

$- 1$ .

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

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

Step 5.1.2.2

Multiply

4

$4$ by

- 1

$- 1$ .

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

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

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

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

Step 5.1.3

Factor

4

$4$ out of

4 - 4 x

$4 - 4 x$ .

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Step 5.1.3.1

Factor

4

$4$ out of

4

$4$ .

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

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

Step 5.1.3.2

Factor

4

$4$ out of

- 4 x

$- 4 x$ .

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

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

Step 5.1.3.3

Factor

4

$4$ out of

4 (1) + 4 (- x)

$4 (1) + 4 (- x)$ .

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

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

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

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

Step 5.1.4

Rewrite

4 (1 - x)

$4 (1 - x)$ as

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

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

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Step 5.1.4.1

Rewrite

4

$4$ as

2^{2}

$2^{2}$ .

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

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

Step 5.1.4.2

Rewrite

1

$1$ as

1^{2}

$1^{2}$ .

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

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

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

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

Step 5.1.5

Pull terms out from under the radical.

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

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

Step 5.1.6

One to any power is one.

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

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

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

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

Step 5.2

Multiply

2

$2$ by

- 1

$- 1$ .

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

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

Step 5.3

Simplify

\frac{- 2 \pm 2 \sqrt{1 - x}}{- 2}

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

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

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

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

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

Step 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}

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