Solve 3(sqrt{2}-2sqrt{3})^2

$$3 ( \sqrt { 2 } - 2 \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$42-12\sqrt{6}\approx 12.606123087$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$3\left(2-4\sqrt{6}+12\right)$$

Add 2 and 12 to get 14.

$$3\left(14-4\sqrt{6}\right)$$

Use the distributive property to multiply 3 by 14-4\sqrt{6}.

$$42-12\sqrt{6}$$

Expand

$$42-12\sqrt{6}$$

Solution Steps

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

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

The square of \sqrt{2} is 2.

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

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

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

The square of \sqrt{3} is 3.

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

Multiply 4 and 3 to get 12.

$$3\left(2-4\sqrt{6}+12\right)$$

Add 2 and 12 to get 14.

$$3\left(14-4\sqrt{6}\right)$$

Use the distributive property to multiply 3 by 14-4\sqrt{6}.

$$42-12\sqrt{6}$$