Simplify ( square root of 8- square root of 3)/( square root of 6-2)

$\frac{\sqrt{8} - \sqrt{3}}{\sqrt{6} - 2}$

Step 1

Simplify the numerator.

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

Factor $4$ out of $8$ .

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

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

$\frac{\sqrt{2^{2} \cdot 2} - \sqrt{3}}{\sqrt{6} - 2}$

Pull terms out from under the radical.

$\frac{2 \sqrt{2} - \sqrt{3}}{\sqrt{6} - 2}$

Step 2

Multiply $\frac{2 \sqrt{2} - \sqrt{3}}{\sqrt{6} - 2}$ by $\frac{\sqrt{6} + 2}{\sqrt{6} + 2}$ .

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

Step 3

Combine fractions.

Multiply $\frac{2 \sqrt{2} - \sqrt{3}}{\sqrt{6} - 2}$ by $\frac{\sqrt{6} + 2}{\sqrt{6} + 2}$ .

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

Expand the denominator using the FOIL method.

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

Simplify.

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

Step 4

Expand $(2 \sqrt{2} - \sqrt{3}) (\sqrt{6} + 2)$ using the FOIL Method.

Apply the distributive property.

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

Apply the distributive property.

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

Apply the distributive property.

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

Step 5

Simplify and combine like terms.

Simplify each term.

Multiply $2 \sqrt{2} \sqrt{6}$ .

Combine using the product rule for radicals.

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

Multiply $6$ by $2$ .

$\frac{2 \sqrt{12} + 2 \sqrt{2} \cdot 2 - \sqrt{3} \sqrt{6} - \sqrt{3} \cdot 2}{2}$

Rewrite $12$ as $2^{2} \cdot 3$ .

Factor $4$ out of $12$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $2$ by $2$ .

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

Multiply $2$ by $2$ .

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

Multiply $- \sqrt{3} \sqrt{6}$ .

Combine using the product rule for radicals.

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

Multiply $6$ by $3$ .

$\frac{4 \sqrt{3} + 4 \sqrt{2} - \sqrt{18} - \sqrt{3} \cdot 2}{2}$

Rewrite $18$ as $3^{2} \cdot 2$ .

Factor $9$ out of $18$ .

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

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

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

Pull terms out from under the radical.

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

Multiply $3$ by $- 1$ .

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

Multiply $2$ by $- 1$ .

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

Subtract $2 \sqrt{3}$ from $4 \sqrt{3}$ .

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

Subtract $3 \sqrt{2}$ from $4 \sqrt{2}$ .

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

Step 6

The result can be shown in multiple forms.

Exact Form:

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

Decimal Form:

$2.43915758 \dots$

Simplify ( square root of 8- square root of 3)/( square root of 6-2)

Related Questions

Support LCMGCF.com by disabling your adblocker.