Solve frac{sqrt{32}-sqrt{18}}{sqrt{8}}

$$\frac{ \sqrt{ 32 } - \sqrt{ 18 } }{ \sqrt{ 8 } }$$

Evaluate

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

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}.

$$\frac{4\sqrt{2}-\sqrt{18}}{\sqrt{8}}$$

Factor 18=3^{2}\times 2. Rewrite the square root of the product \sqrt{3^{2}\times 2} as the product of square roots \sqrt{3^{2}}\sqrt{2}. Take the square root of 3^{2}.

$$\frac{4\sqrt{2}-3\sqrt{2}}{\sqrt{8}}$$

Combine 4\sqrt{2} and -3\sqrt{2} to get \sqrt{2}.

$$\frac{\sqrt{2}}{\sqrt{8}}$$

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}.

$$\frac{\sqrt{2}}{2\sqrt{2}}$$

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

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

The square of \sqrt{2} is 2.

$$\frac{\sqrt{2}\sqrt{2}}{2\times 2}$$

Multiply \sqrt{2} and \sqrt{2} to get 2.

$$\frac{2}{2\times 2}$$

Multiply 2 and 2 to get 4.

$$\frac{2}{4}$$

Reduce the fraction \frac{2}{4} to lowest terms by extracting and canceling out 2.

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

Factor

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