Solve 2(2sqrt{3}-4sqrt{2})^2

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

Evaluate

$$88-32\sqrt{6}\approx 9.616328231$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-4\sqrt{2}\right)^{2}.

$$2\left(4\left(\sqrt{3}\right)^{2}-16\sqrt{3}\sqrt{2}+16\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$2\left(4\times 3-16\sqrt{3}\sqrt{2}+16\left(\sqrt{2}\right)^{2}\right)$$

Multiply 4 and 3 to get 12.

$$2\left(12-16\sqrt{3}\sqrt{2}+16\left(\sqrt{2}\right)^{2}\right)$$

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

$$2\left(12-16\sqrt{6}+16\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$2\left(12-16\sqrt{6}+16\times 2\right)$$

Multiply 16 and 2 to get 32.

$$2\left(12-16\sqrt{6}+32\right)$$

Add 12 and 32 to get 44.

$$2\left(44-16\sqrt{6}\right)$$

Use the distributive property to multiply 2 by 44-16\sqrt{6}.

$$88-32\sqrt{6}$$

Expand

$$88-32\sqrt{6}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-4\sqrt{2}\right)^{2}.

$$2\left(4\left(\sqrt{3}\right)^{2}-16\sqrt{3}\sqrt{2}+16\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{3} is 3.

$$2\left(4\times 3-16\sqrt{3}\sqrt{2}+16\left(\sqrt{2}\right)^{2}\right)$$

Multiply 4 and 3 to get 12.

$$2\left(12-16\sqrt{3}\sqrt{2}+16\left(\sqrt{2}\right)^{2}\right)$$

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

$$2\left(12-16\sqrt{6}+16\left(\sqrt{2}\right)^{2}\right)$$

The square of \sqrt{2} is 2.

$$2\left(12-16\sqrt{6}+16\times 2\right)$$

Multiply 16 and 2 to get 32.

$$2\left(12-16\sqrt{6}+32\right)$$

Add 12 and 32 to get 44.

$$2\left(44-16\sqrt{6}\right)$$

Use the distributive property to multiply 2 by 44-16\sqrt{6}.

$$88-32\sqrt{6}$$