Solve frac{sqrt{32}+sqrt{48}}{sqrt{8}+sqrt{12}}

$$\frac { \sqrt { 32 } + \sqrt { 48 } } { \sqrt { 8 } + \sqrt { 12 } }$$

Evaluate

$$2$$

Solution Steps

Factor 32=4^{2}\times 2. Rewrite the square root of the product \sqrt{4^{2}\times 2} as the product of square roots \sqrt{4^{2}}\sqrt{2}. Take the square root of 4^{2}.

$$\frac{4\sqrt{2}+\sqrt{48}}{\sqrt{8}+\sqrt{12}}$$

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

$$\frac{4\sqrt{2}+4\sqrt{3}}{\sqrt{8}+\sqrt{12}}$$

Factor 8=2^{2}\times 2. Rewrite the square root of the product \sqrt{2^{2}\times 2} as the product of square roots \sqrt{2^{2}}\sqrt{2}. Take the square root of 2^{2}.

$$\frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+\sqrt{12}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

Rationalize the denominator of \frac{4\sqrt{2}+4\sqrt{3}}{2\sqrt{2}+2\sqrt{3}} by multiplying numerator and denominator by 2\sqrt{2}-2\sqrt{3}.

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

Consider \left(2\sqrt{2}+2\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Expand \left(2\sqrt{2}\right)^{2}.

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

Calculate 2 to the power of 2 and get 4.

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

The square of \sqrt{2} is 2.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{4\times 2-\left(2\sqrt{3}\right)^{2}}$$

Multiply 4 and 2 to get 8.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-\left(2\sqrt{3}\right)^{2}}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-2^{2}\left(\sqrt{3}\right)^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-4\times 3}$$

Multiply 4 and 3 to get 12.

$$\frac{\left(4\sqrt{2}+4\sqrt{3}\right)\left(2\sqrt{2}-2\sqrt{3}\right)}{8-12}$$

Subtract 12 from 8 to get -4.

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

Apply the distributive property by multiplying each term of 4\sqrt{2}+4\sqrt{3} by each term of 2\sqrt{2}-2\sqrt{3}.

$$\frac{8\left(\sqrt{2}\right)^{2}-8\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}$$

The square of \sqrt{2} is 2.

$$\frac{8\times 2-8\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}$$

Multiply 8 and 2 to get 16.

$$\frac{16-8\sqrt{3}\sqrt{2}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{16-8\sqrt{6}+8\sqrt{3}\sqrt{2}-8\left(\sqrt{3}\right)^{2}}{-4}$$

To multiply \sqrt{3} and \sqrt{2}, multiply the numbers under the square root.

$$\frac{16-8\sqrt{6}+8\sqrt{6}-8\left(\sqrt{3}\right)^{2}}{-4}$$

Combine -8\sqrt{6} and 8\sqrt{6} to get 0.

$$\frac{16-8\left(\sqrt{3}\right)^{2}}{-4}$$

The square of \sqrt{3} is 3.

$$\frac{16-8\times 3}{-4}$$

Multiply -8 and 3 to get -24.

$$\frac{16-24}{-4}$$

Subtract 24 from 16 to get -8.

$$\frac{-8}{-4}$$

Divide -8 by -4 to get 2.

$$2$$

Factor

$$2$$