Solve frac{sqrt{128}+3sqrt{8}}{7sqrt{32}}

$$\frac { \sqrt { 128 } + 3 \sqrt { 8 } } { 7 \sqrt { 32 } }$$

Evaluate

$$\frac{1}{2}=0.5$$

Solution Steps

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}.

$$\frac{8\sqrt{2}+3\sqrt{8}}{7\sqrt{32}}$$

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}.

$$\frac{8\sqrt{2}+3\times 2\sqrt{2}}{7\sqrt{32}}$$

Multiply 3 and 2 to get 6.

$$\frac{8\sqrt{2}+6\sqrt{2}}{7\sqrt{32}}$$

Combine 8\sqrt{2} and 6\sqrt{2} to get 14\sqrt{2}.

$$\frac{14\sqrt{2}}{7\sqrt{32}}$$

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}.

$$\frac{14\sqrt{2}}{7\times 4\sqrt{2}}$$

Multiply 7 and 4 to get 28.

$$\frac{14\sqrt{2}}{28\sqrt{2}}$$

Cancel out 14\sqrt{2} in both numerator and denominator.

$$\frac{1}{2}$$

Factor

$$\frac{1}{2} = 0.5$$