Linear Algebra Examples

$\frac{- 32 + \sqrt{512}}{- 32}$

Step 1

Simplify the numerator.

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Step 1.1

Rewrite $512$ as $16^{2} \cdot 2$ .

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Step 1.1.1

Factor $256$ out of $512$ .

$\frac{- 32 + \sqrt{256 (2)}}{- 32}$

Step 1.1.2

Rewrite $256$ as $16^{2}$ .

$\frac{- 32 + \sqrt{16^{2} \cdot 2}}{- 32}$

$\frac{- 32 + \sqrt{16^{2} \cdot 2}}{- 32}$

Step 1.2

Pull terms out from under the radical.

$\frac{- 32 + 16 \sqrt{2}}{- 32}$

$\frac{- 32 + 16 \sqrt{2}}{- 32}$

Step 2

Simplify terms.

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Step 2.1

Cancel the common factor of $- 32 + 16 \sqrt{2}$ and $- 32$ .

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Step 2.1.1

Factor $16$ out of $- 32$ .

$\frac{16 \cdot - 2 + 16 \sqrt{2}}{- 32}$

Step 2.1.2

Factor $16$ out of $16 \sqrt{2}$ .

$\frac{16 \cdot - 2 + 16 (\sqrt{2})}{- 32}$

Step 2.1.3

Factor $16$ out of $16 \cdot - 2 + 16 (\sqrt{2})$ .

$\frac{16 \cdot (- 2 + \sqrt{2})}{- 32}$

Step 2.1.4

Cancel the common factors.

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Step 2.1.4.1

Factor $16$ out of $- 32$ .

$\frac{16 \cdot (- 2 + \sqrt{2})}{16 (- 2)}$

Step 2.1.4.2

Cancel the common factor.

$\frac{16 \cdot (- 2 + \sqrt{2})}{16 \cdot - 2}$

Step 2.1.4.3

Rewrite the expression.

$\frac{- 2 + \sqrt{2}}{- 2}$

$\frac{- 2 + \sqrt{2}}{- 2}$

$\frac{- 2 + \sqrt{2}}{- 2}$

Step 2.2

Move the negative in front of the fraction.

$- \frac{- 2 + \sqrt{2}}{2}$

Step 2.3

Rewrite $- 2$ as $- 1 (2)$ .

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

Step 2.4

Factor $- 1$ out of $\sqrt{2}$ .

$- \frac{- 1 (2) - 1 (- \sqrt{2})}{2}$

Step 2.5

Factor $- 1$ out of $- 1 (2) - 1 (- \sqrt{2})$ .

$- \frac{- 1 (2 - \sqrt{2})}{2}$

Step 2.6

Simplify the expression.

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Step 2.6.1

Move the negative in front of the fraction.

$- - \frac{2 - \sqrt{2}}{2}$

Step 2.6.2

Multiply $- 1$ by $- 1$ .

$1 \frac{2 - \sqrt{2}}{2}$

Step 2.6.3

Multiply $\frac{2 - \sqrt{2}}{2}$ by $1$ .

$\frac{2 - \sqrt{2}}{2}$

$\frac{2 - \sqrt{2}}{2}$

$\frac{2 - \sqrt{2}}{2}$

Step 3

The result can be shown in multiple forms.

Exact Form:

$\frac{2 - \sqrt{2}}{2}$

Decimal Form:

$0.29289321 \dots$

$[\begin{array}{c} x^{2} & \frac{1}{2} \\ \sqrt{π} & \int x d x \end{array}]$