Solve sqrt{18}*sqrt{2/3}-(1-sqrt{3})^2

$$\sqrt { 18 } \times \sqrt { \frac { 2 } { 3 } } - ( 1 - \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$4\left(\sqrt{3}-1\right)\approx 2.92820323$$

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}.

$$3\sqrt{2}\sqrt{\frac{2}{3}}-\left(1-\sqrt{3}\right)^{2}$$

Rewrite the square root of the division \sqrt{\frac{2}{3}} as the division of square roots \frac{\sqrt{2}}{\sqrt{3}}.

$$3\sqrt{2}\times \frac{\sqrt{2}}{\sqrt{3}}-\left(1-\sqrt{3}\right)^{2}$$

Rationalize the denominator of \frac{\sqrt{2}}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$3\sqrt{2}\times \frac{\sqrt{2}\sqrt{3}}{\left(\sqrt{3}\right)^{2}}-\left(1-\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$3\sqrt{2}\times \frac{\sqrt{2}\sqrt{3}}{3}-\left(1-\sqrt{3}\right)^{2}$$

To multiply \sqrt{2} and \sqrt{3}, multiply the numbers under the square root.

$$3\sqrt{2}\times \frac{\sqrt{6}}{3}-\left(1-\sqrt{3}\right)^{2}$$

Cancel out 3 and 3.

$$\sqrt{6}\sqrt{2}-\left(1-\sqrt{3}\right)^{2}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(1-\sqrt{3}\right)^{2}.

$$\sqrt{6}\sqrt{2}-\left(1-2\sqrt{3}+\left(\sqrt{3}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$\sqrt{6}\sqrt{2}-\left(1-2\sqrt{3}+3\right)$$

Add 1 and 3 to get 4.

$$\sqrt{6}\sqrt{2}-\left(4-2\sqrt{3}\right)$$

To find the opposite of 4-2\sqrt{3}, find the opposite of each term.

$$\sqrt{6}\sqrt{2}-4+2\sqrt{3}$$

Factor 6=2\times 3. Rewrite the square root of the product \sqrt{2\times 3} as the product of square roots \sqrt{2}\sqrt{3}.

$$\sqrt{2}\sqrt{3}\sqrt{2}-4+2\sqrt{3}$$

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

$$2\sqrt{3}-4+2\sqrt{3}$$

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

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