Calculus Examples

$\frac{1}{4} y^{2} + x = 0$

Step 1

Solve for $y$ .

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Step 1.1

Combine $\frac{1}{4}$ and $y^{2}$ .

$\frac{y^{2}}{4} + x = 0$

Step 1.2

Subtract $x$ from both sides of the equation.

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

Step 1.3

Multiply both sides of the equation by $4$ .

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

Step 1.4

Simplify both sides of the equation.

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Step 1.4.1

Simplify the left side.

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Step 1.4.1.1

Cancel the common factor of $4$ .

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Step 1.4.1.1.1

Cancel the common factor.

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

Step 1.4.1.1.2

Rewrite the expression.

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

Step 1.4.2

Simplify the right side.

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Step 1.4.2.1

Multiply $- 1$ by $4$ .

$y^{2} = - 4 x$

Step 1.5

Take the specified root of both sides of the equation to eliminate the exponent on the left side.

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

Step 1.6

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

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Step 1.6.1

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

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Step 1.6.1.1

Factor $4$ out of $- 4$ .

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

Step 1.6.1.2

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

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

Step 1.6.1.3

Add parentheses.

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

Step 1.6.2

Pull terms out from under the radical.

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

Step 1.7

The complete solution is the result of both the positive and negative portions of the solution.

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Step 1.7.1

First, use the positive value of the $\pm$ to find the first solution.

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

Step 1.7.2

Next, use the negative value of the $\pm$ to find the second solution.

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

Step 1.7.3

The complete solution is the result of both the positive and negative portions of the solution.

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

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

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

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

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

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

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

Rewrite $- 1 x$ as $- x$ .

$y = 2 \sqrt{- x}$

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

Step 4