Solve 12-{left(3sqrt{2}-sqrt{6}right)}^2

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

Evaluate

$$12\left(\sqrt{3}-1\right)\approx 8.784609691$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-\sqrt{6}\right)^{2}.

$$12-\left(9\left(\sqrt{2}\right)^{2}-6\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$12-\left(9\times 2-6\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)$$

Multiply 9 and 2 to get 18.

$$12-\left(18-6\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)$$

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}.

$$12-\left(18-6\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}\right)$$

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

$$12-\left(18-6\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}\right)$$

Multiply -6 and 2 to get -12.

$$12-\left(18-12\sqrt{3}+\left(\sqrt{6}\right)^{2}\right)$$

The square of \sqrt{6} is 6.

$$12-\left(18-12\sqrt{3}+6\right)$$

Add 18 and 6 to get 24.

$$12-\left(24-12\sqrt{3}\right)$$

To find the opposite of 24-12\sqrt{3}, find the opposite of each term.

$$12-24+12\sqrt{3}$$

Subtract 24 from 12 to get -12.

$$-12+12\sqrt{3}$$

Expand

$$12 \sqrt{3} - 12 = 8.784609691$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(3\sqrt{2}-\sqrt{6}\right)^{2}.

$$12-\left(9\left(\sqrt{2}\right)^{2}-6\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$12-\left(9\times 2-6\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)$$

Multiply 9 and 2 to get 18.

$$12-\left(18-6\sqrt{2}\sqrt{6}+\left(\sqrt{6}\right)^{2}\right)$$

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}.

$$12-\left(18-6\sqrt{2}\sqrt{2}\sqrt{3}+\left(\sqrt{6}\right)^{2}\right)$$

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

$$12-\left(18-6\times 2\sqrt{3}+\left(\sqrt{6}\right)^{2}\right)$$

Multiply -6 and 2 to get -12.

$$12-\left(18-12\sqrt{3}+\left(\sqrt{6}\right)^{2}\right)$$

The square of \sqrt{6} is 6.

$$12-\left(18-12\sqrt{3}+6\right)$$

Add 18 and 6 to get 24.

$$12-\left(24-12\sqrt{3}\right)$$

To find the opposite of 24-12\sqrt{3}, find the opposite of each term.

$$12-24+12\sqrt{3}$$

Subtract 24 from 12 to get -12.

$$-12+12\sqrt{3}$$