Solve frac{2sqrt{32}+4sqrt{8}}{2sqrt{72}-sqrt{128}}

$$\frac { 2 \sqrt { 32 } + 4 \sqrt { 8 } } { 2 \sqrt { 72 } - \sqrt { 128 } }$$

Evaluate

$$4$$

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}.

$$\frac{2\times 4\sqrt{2}+4\sqrt{8}}{2\sqrt{72}-\sqrt{128}}$$

Multiply 2 and 4 to get 8.

$$\frac{8\sqrt{2}+4\sqrt{8}}{2\sqrt{72}-\sqrt{128}}$$

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}.

$$\frac{8\sqrt{2}+4\times 2\sqrt{2}}{2\sqrt{72}-\sqrt{128}}$$

Multiply 4 and 2 to get 8.

$$\frac{8\sqrt{2}+8\sqrt{2}}{2\sqrt{72}-\sqrt{128}}$$

Combine 8\sqrt{2} and 8\sqrt{2} to get 16\sqrt{2}.

$$\frac{16\sqrt{2}}{2\sqrt{72}-\sqrt{128}}$$

Factor 72=6^{2}\times 2. Rewrite the square root of the product \sqrt{6^{2}\times 2} as the product of square roots \sqrt{6^{2}}\sqrt{2}. Take the square root of 6^{2}.

$$\frac{16\sqrt{2}}{2\times 6\sqrt{2}-\sqrt{128}}$$

Multiply 2 and 6 to get 12.

$$\frac{16\sqrt{2}}{12\sqrt{2}-\sqrt{128}}$$

Factor 128=8^{2}\times 2. Rewrite the square root of the product \sqrt{8^{2}\times 2} as the product of square roots \sqrt{8^{2}}\sqrt{2}. Take the square root of 8^{2}.

$$\frac{16\sqrt{2}}{12\sqrt{2}-8\sqrt{2}}$$

Combine 12\sqrt{2} and -8\sqrt{2} to get 4\sqrt{2}.

$$\frac{16\sqrt{2}}{4\sqrt{2}}$$

Cancel out 4\sqrt{2} in both numerator and denominator.

$$4$$

Factor

$$2^{2}$$