Solve (4sqrt{3}-2sqrt{6})^2

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

Evaluate

$$72-48\sqrt{2}\approx 4.117749006$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{3}-2\sqrt{6}\right)^{2}.

$$16\left(\sqrt{3}\right)^{2}-16\sqrt{3}\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{3} is 3.

$$16\times 3-16\sqrt{3}\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

Multiply 16 and 3 to get 48.

$$48-16\sqrt{3}\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

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

$$48-16\sqrt{3}\sqrt{3}\sqrt{2}+4\left(\sqrt{6}\right)^{2}$$

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

$$48-16\times 3\sqrt{2}+4\left(\sqrt{6}\right)^{2}$$

Multiply -16 and 3 to get -48.

$$48-48\sqrt{2}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$48-48\sqrt{2}+4\times 6$$

Multiply 4 and 6 to get 24.

$$48-48\sqrt{2}+24$$

Add 48 and 24 to get 72.

$$72-48\sqrt{2}$$

Expand

$$72-48\sqrt{2}$$

Solution Steps

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(4\sqrt{3}-2\sqrt{6}\right)^{2}.

$$16\left(\sqrt{3}\right)^{2}-16\sqrt{3}\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{3} is 3.

$$16\times 3-16\sqrt{3}\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

Multiply 16 and 3 to get 48.

$$48-16\sqrt{3}\sqrt{6}+4\left(\sqrt{6}\right)^{2}$$

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

$$48-16\sqrt{3}\sqrt{3}\sqrt{2}+4\left(\sqrt{6}\right)^{2}$$

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

$$48-16\times 3\sqrt{2}+4\left(\sqrt{6}\right)^{2}$$

Multiply -16 and 3 to get -48.

$$48-48\sqrt{2}+4\left(\sqrt{6}\right)^{2}$$

The square of \sqrt{6} is 6.

$$48-48\sqrt{2}+4\times 6$$

Multiply 4 and 6 to get 24.

$$48-48\sqrt{2}+24$$

Add 48 and 24 to get 72.

$$72-48\sqrt{2}$$