Solve sqrt{45}+sqrt{5}-3sqrt{1/3}*sqrt{27}

$$\sqrt { 45 } + \sqrt { 5 } - 3 \sqrt { \frac { 1 } { 3 } } \times \sqrt { 27 }$$

Evaluate

$$4\sqrt{5}-9\approx -0.05572809$$

Solution Steps

Factor 45=3^{2}\times 5. Rewrite the square root of the product \sqrt{3^{2}\times 5} as the product of square roots \sqrt{3^{2}}\sqrt{5}. Take the square root of 3^{2}.

$$3\sqrt{5}+\sqrt{5}-3\sqrt{\frac{1}{3}}\sqrt{27}$$

Combine 3\sqrt{5} and \sqrt{5} to get 4\sqrt{5}.

$$4\sqrt{5}-3\sqrt{\frac{1}{3}}\sqrt{27}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$4\sqrt{5}-3\times \frac{\sqrt{1}}{\sqrt{3}}\sqrt{27}$$

Calculate the square root of 1 and get 1.

$$4\sqrt{5}-3\times \frac{1}{\sqrt{3}}\sqrt{27}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$4\sqrt{5}-3\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}\sqrt{27}$$

The square of \sqrt{3} is 3.

$$4\sqrt{5}-3\times \frac{\sqrt{3}}{3}\sqrt{27}$$

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

$$4\sqrt{5}-3\times \frac{\sqrt{3}}{3}\times 3\sqrt{3}$$

Multiply 3 and 3 to get 9.

$$4\sqrt{5}-9\times \frac{\sqrt{3}}{3}\sqrt{3}$$

Cancel out 3, the greatest common factor in 9 and 3.

$$4\sqrt{5}-3\sqrt{3}\sqrt{3}$$

Multiply \sqrt{3} and \sqrt{3} to get 3.

$$4\sqrt{5}-3\times 3$$

Multiply 3 and 3 to get 9.

$$4\sqrt{5}-9$$