Algebra Examples

$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{8}}$

Step 1

Simplify the denominator.

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Step 1.1

Rewrite $8$ as $2^{2} \cdot 2$ .

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Step 1.1.1

Factor $4$ out of $8$ .

$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{4 (2)}}$

Step 1.1.2

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

$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - \sqrt{2^{2} \cdot 2}}$

Step 1.2

Pull terms out from under the radical.

$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - (2 \sqrt{2})}$

Step 1.3

Multiply $2$ by $- 1$ .

$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - 2 \sqrt{2}}$

Step 2

Multiply $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - 2 \sqrt{2}}$ by $\frac{\sqrt{3} + 2 \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}$ .

$\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - 2 \sqrt{2}} \cdot \frac{\sqrt{3} + 2 \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}$

Step 3

Combine fractions.

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Step 3.1

Multiply $\frac{\sqrt{2} + \sqrt{3}}{\sqrt{3} - 2 \sqrt{2}}$ by $\frac{\sqrt{3} + 2 \sqrt{2}}{\sqrt{3} + 2 \sqrt{2}}$ .

$\frac{(\sqrt{2} + \sqrt{3}) (\sqrt{3} + 2 \sqrt{2})}{(\sqrt{3} - 2 \sqrt{2}) (\sqrt{3} + 2 \sqrt{2})}$

Step 3.2

Expand the denominator using the FOIL method.

$\frac{(\sqrt{2} + \sqrt{3}) (\sqrt{3} + 2 \sqrt{2})}{{\sqrt{3}}^{2} + 2 \sqrt{6} - 2 \sqrt{6} - 4 {\sqrt{2}}^{2}}$

Step 3.3

Simplify.

$\frac{(\sqrt{2} + \sqrt{3}) (\sqrt{3} + 2 \sqrt{2})}{- 5}$

Step 4

Expand $(\sqrt{2} + \sqrt{3}) (\sqrt{3} + 2 \sqrt{2})$ using the FOIL Method.

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Step 4.1

Apply the distributive property.

$\frac{\sqrt{2} (\sqrt{3} + 2 \sqrt{2}) + \sqrt{3} (\sqrt{3} + 2 \sqrt{2})}{- 5}$

Step 4.2

Apply the distributive property.

$\frac{\sqrt{2} \sqrt{3} + \sqrt{2} (2 \sqrt{2}) + \sqrt{3} (\sqrt{3} + 2 \sqrt{2})}{- 5}$

Step 4.3

Apply the distributive property.

$\frac{\sqrt{2} \sqrt{3} + \sqrt{2} (2 \sqrt{2}) + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5

Simplify and combine like terms.

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Step 5.1

Simplify each term.

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Step 5.1.1

Combine using the product rule for radicals.

$\frac{\sqrt{2 \cdot 3} + \sqrt{2} (2 \sqrt{2}) + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.2

Multiply $2$ by $3$ .

$\frac{\sqrt{6} + \sqrt{2} (2 \sqrt{2}) + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.3

Multiply $\sqrt{2} (2 \sqrt{2})$ .

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Step 5.1.3.1

Raise $\sqrt{2}$ to the power of $1$ .

$\frac{\sqrt{6} + 2 ({\sqrt{2}}^{1} \sqrt{2}) + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.3.2

Raise $\sqrt{2}$ to the power of $1$ .

$\frac{\sqrt{6} + 2 ({\sqrt{2}}^{1} {\sqrt{2}}^{1}) + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.3.3

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

$\frac{\sqrt{6} + 2 {\sqrt{2}}^{1 + 1} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.3.4

Add $1$ and $1$ .

$\frac{\sqrt{6} + 2 {\sqrt{2}}^{2} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.4

Rewrite ${\sqrt{2}}^{2}$ as $2$ .

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Step 5.1.4.1

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

$\frac{\sqrt{6} + 2 {(2^{\frac{1}{2}})}^{2} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.4.2

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

$\frac{\sqrt{6} + 2 \cdot 2^{\frac{1}{2} \cdot 2} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.4.3

Combine $\frac{1}{2}$ and $2$ .

$\frac{\sqrt{6} + 2 \cdot 2^{\frac{2}{2}} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.4.4

Cancel the common factor of $2$ .

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Step 5.1.4.4.1

Cancel the common factor.

$\frac{\sqrt{6} + 2 \cdot 2^{\frac{2}{2}} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.4.4.2

Rewrite the expression.

$\frac{\sqrt{6} + 2 \cdot 2^{1} + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.4.5

Evaluate the exponent.

$\frac{\sqrt{6} + 2 \cdot 2 + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.5

Multiply $2$ by $2$ .

$\frac{\sqrt{6} + 4 + \sqrt{3} \sqrt{3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.6

Combine using the product rule for radicals.

$\frac{\sqrt{6} + 4 + \sqrt{3 \cdot 3} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.7

Multiply $3$ by $3$ .

$\frac{\sqrt{6} + 4 + \sqrt{9} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.8

Rewrite $9$ as $3^{2}$ .

$\frac{\sqrt{6} + 4 + \sqrt{3^{2}} + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.9

Pull terms out from under the radical, assuming positive real numbers.

$\frac{\sqrt{6} + 4 + 3 + \sqrt{3} (2 \sqrt{2})}{- 5}$

Step 5.1.10

Multiply $\sqrt{3} (2 \sqrt{2})$ .

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Step 5.1.10.1

Combine using the product rule for radicals.

$\frac{\sqrt{6} + 4 + 3 + 2 \sqrt{3 \cdot 2}}{- 5}$

Step 5.1.10.2

Multiply $3$ by $2$ .

$\frac{\sqrt{6} + 4 + 3 + 2 \sqrt{6}}{- 5}$

Step 5.2

Add $\sqrt{6}$ and $2 \sqrt{6}$ .

$\frac{4 + 3 + 3 \sqrt{6}}{- 5}$

Step 5.3

Add $4$ and $3$ .

$\frac{7 + 3 \sqrt{6}}{- 5}$

Step 6

Move the negative in front of the fraction.

$- \frac{7 + 3 \sqrt{6}}{5}$

Step 7

The result can be shown in multiple forms.

Exact Form:

$- \frac{7 + 3 \sqrt{6}}{5}$

Decimal Form:

$- 2.86969384 \dots$