Solve frac{sqrt{72}+sqrt{32}-4}{sqrt{8}}=

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

Evaluate

$$5-\sqrt{2}\approx 3.585786438$$

Solution Steps

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}.

$$\frac{6\sqrt{2}+\sqrt{32}-4}{\sqrt{8}}$$

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}.

$$\frac{6\sqrt{2}+4\sqrt{2}-4}{\sqrt{8}}$$

Combine 6\sqrt{2} and 4\sqrt{2} to get 10\sqrt{2}.

$$\frac{10\sqrt{2}-4}{\sqrt{8}}$$

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}.

$$\frac{10\sqrt{2}-4}{2\sqrt{2}}$$

Rationalize the denominator of \frac{10\sqrt{2}-4}{2\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

$$\frac{\left(10\sqrt{2}-4\right)\sqrt{2}}{2\left(\sqrt{2}\right)^{2}}$$

The square of \sqrt{2} is 2.

$$\frac{\left(10\sqrt{2}-4\right)\sqrt{2}}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$\frac{\left(10\sqrt{2}-4\right)\sqrt{2}}{4}$$

Use the distributive property to multiply 10\sqrt{2}-4 by \sqrt{2}.

$$\frac{10\left(\sqrt{2}\right)^{2}-4\sqrt{2}}{4}$$

The square of \sqrt{2} is 2.

$$\frac{10\times 2-4\sqrt{2}}{4}$$

Multiply 10 and 2 to get 20.

$$\frac{20-4\sqrt{2}}{4}$$

Divide each term of 20-4\sqrt{2} by 4 to get 5-\sqrt{2}.

$$5-\sqrt{2}$$