Calculus Examples

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

Step 1

Rewrite the equation as $\frac{1}{2 \sqrt{2 x - x^{2}}} = y$ .

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

Step 2

Cross multiply.

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Step 2.1

Cross multiply by setting the product of the numerator of the right side and the denominator of the left side equal to the product of the numerator of the left side and the denominator of the right side.

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

Step 2.2

Simplify the left side.

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Step 2.2.1

Simplify $y \cdot (2 \sqrt{2 x - x^{2}})$ .

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Step 2.2.1.1

Rewrite using the commutative property of multiplication.

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

Step 2.2.1.2

Factor $x$ out of $2 x - x^{2}$ .

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Step 2.2.1.2.1

Factor $x$ out of $2 x$ .

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

Step 2.2.1.2.2

Factor $x$ out of $- x^{2}$ .

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

Step 2.2.1.2.3

Factor $x$ out of $x \cdot 2 + x (- x)$ .

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

Step 3

To remove the radical on the left side of the equation, square both sides of the equation.

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

Step 4

Simplify each side of the equation.

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Step 4.1

Use $\sqrt[n]{a^{x}} = a^{\frac{x}{n}}$ to rewrite $\sqrt{x (2 - x)}$ as ${(x (2 - x))}^{\frac{1}{2}}$ .

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

Step 4.2

Simplify the left side.

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Step 4.2.1

Simplify ${(2 y {(x (2 - x))}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.1.1

Simplify by multiplying through.

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Step 4.2.1.1.1

Apply the distributive property.

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

Step 4.2.1.1.2

Reorder.

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Step 4.2.1.1.2.1

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

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

Step 4.2.1.1.2.2

Rewrite using the commutative property of multiplication.

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

Step 4.2.1.2

Multiply $x$ by $x$ by adding the exponents.

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Step 4.2.1.2.1

Move $x$ .

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

Step 4.2.1.2.2

Multiply $x$ by $x$ .

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

Step 4.2.1.3

Use the power rule ${(a b)}^{n} = a^{n} b^{n}$ to distribute the exponent.

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Step 4.2.1.3.1

Apply the product rule to $2 y {(2 x - x^{2})}^{\frac{1}{2}}$ .

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

Step 4.2.1.3.2

Apply the product rule to $2 y$ .

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

Step 4.2.1.4

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

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

Step 4.2.1.5

Multiply the exponents in ${({(2 x - x^{2})}^{\frac{1}{2}})}^{2}$ .

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Step 4.2.1.5.1

Apply the power rule and multiply exponents, ${(a^{m})}^{n} = a^{m n}$ .

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

Step 4.2.1.5.2

Cancel the common factor of $2$ .

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Step 4.2.1.5.2.1

Cancel the common factor.

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

Step 4.2.1.5.2.2

Rewrite the expression.

$4 y^{2} {(2 x - x^{2})}^{1} = 1^{2}$

Step 4.2.1.6

Simplify.

$4 y^{2} (2 x - x^{2}) = 1^{2}$

Step 4.2.1.7

Simplify by multiplying through.

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Step 4.2.1.7.1

Apply the distributive property.

$4 y^{2} (2 x) + 4 y^{2} (- x^{2}) = 1^{2}$

Step 4.2.1.7.2

Reorder.

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Step 4.2.1.7.2.1

Rewrite using the commutative property of multiplication.

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

Step 4.2.1.7.2.2

Rewrite using the commutative property of multiplication.

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

Step 4.2.1.8

Simplify each term.

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Step 4.2.1.8.1

Multiply $4$ by $2$ .

$8 y^{2} x + 4 \cdot - 1 y^{2} x^{2} = 1^{2}$

Step 4.2.1.8.2

Multiply $4$ by $- 1$ .

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

Step 4.3

Simplify the right side.

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Step 4.3.1

One to any power is one.

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

Step 5

Solve for $x$ .

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Step 5.1

Subtract $1$ from both sides of the equation.

$8 y^{2} x - 4 y^{2} x^{2} - 1 = 0$

Step 5.2

Use the quadratic formula to find the solutions.

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

Step 5.3

Substitute the values $a = - 4 y^{2}$ , $b = 8 y^{2}$ , and $c = - 1$ into the quadratic formula and solve for $x$ .

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

Step 5.4

Simplify.

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Step 5.4.1

Simplify the numerator.

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Step 5.4.1.1

Rewrite $- 4 \cdot - 4 y^{2}$ as ${(4 y)}^{2}$ .

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

Step 5.4.1.2

Since both terms are perfect squares, factor using the difference of squares formula, $a^{2} - b^{2} = (a + b) (a - b)$ where $a = 8 y^{2}$ and $b = 4 y$ .

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

Step 5.4.1.3

Simplify.

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Step 5.4.1.3.1

Factor $4 y$ out of $8 y^{2} + 4 y$ .

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Step 5.4.1.3.1.1

Factor $4 y$ out of $8 y^{2}$ .

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

Step 5.4.1.3.1.2

Factor $4 y$ out of $4 y$ .

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

Step 5.4.1.3.1.3

Factor $4 y$ out of $4 y (2 y) + 4 y (1)$ .

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

Step 5.4.1.3.2

Multiply $4$ by $- 1$ .

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

Step 5.4.1.4

Factor $4 y$ out of $8 y^{2} - 4 y$ .

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Step 5.4.1.4.1

Factor $4 y$ out of $8 y^{2}$ .

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

Step 5.4.1.4.2

Factor $4 y$ out of $- 4 y$ .

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

Step 5.4.1.4.3

Factor $4 y$ out of $4 y (2 y) + 4 y (- 1)$ .

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

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

Step 5.4.1.5

Combine exponents.

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Step 5.4.1.5.1

Multiply $4$ by $4$ .

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

Step 5.4.1.5.2

Raise $y$ to the power of $1$ .

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

Step 5.4.1.5.3

Raise $y$ to the power of $1$ .

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

Step 5.4.1.5.4

Use the power rule $a^{m} a^{n} = a^{m + n}$ to combine exponents.

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

Step 5.4.1.5.5

Add $1$ and $1$ .

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

Step 5.4.1.6

Rewrite $16 y^{2} (2 y + 1) (2 y - 1)$ as ${(4 y)}^{2} ((2 y + 1) (2 y - 1))$ .

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Step 5.4.1.6.1

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

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

Step 5.4.1.6.2

Rewrite $4^{2} y^{2}$ as ${(4 y)}^{2}$ .

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

Step 5.4.1.6.3

Add parentheses.

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

Step 5.4.1.7

Pull terms out from under the radical.

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

Step 5.4.2

Multiply $- 4$ by $2$ .

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

Step 5.4.3

Simplify $\frac{- 8 y^{2} \pm 4 y \sqrt{(2 y + 1) (2 y - 1)}}{- 8 y^{2}}$ .

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

Step 5.5

The final answer is the combination of both solutions.

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

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

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

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