Solve sqrt{18}+sqrt{1/2}-sqrt{8}

$$\sqrt { 18 } + \sqrt { \frac { 1 } { 2 } } - \sqrt { 8 }$$

Evaluate

$$\frac{3\sqrt{2}}{2}\approx 2.121320344$$

Solution Steps

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}.

$$3\sqrt{2}+\sqrt{\frac{1}{2}}-\sqrt{8}$$

Rewrite the square root of the division \sqrt{\frac{1}{2}} as the division of square roots \frac{\sqrt{1}}{\sqrt{2}}.

$$3\sqrt{2}+\frac{\sqrt{1}}{\sqrt{2}}-\sqrt{8}$$

Calculate the square root of 1 and get 1.

$$3\sqrt{2}+\frac{1}{\sqrt{2}}-\sqrt{8}$$

Rationalize the denominator of \frac{1}{\sqrt{2}} by multiplying numerator and denominator by \sqrt{2}.

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

The square of \sqrt{2} is 2.

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

Combine 3\sqrt{2} and \frac{\sqrt{2}}{2} to get \frac{7}{2}\sqrt{2}.

$$\frac{7}{2}\sqrt{2}-\sqrt{8}$$

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}.

$$\frac{7}{2}\sqrt{2}-2\sqrt{2}$$

Combine \frac{7}{2}\sqrt{2} and -2\sqrt{2} to get \frac{3}{2}\sqrt{2}.

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