Solve (5)(sqrt{3}-1)^2-(2sqrt{3})^2

$$( 5 ) ( \sqrt { 3 } - 1 ) ^ { 2 } - ( 2 \sqrt { 3 } ) ^ { 2 }$$

Evaluate

$$8-10\sqrt{3}\approx -9.320508076$$

Solution Steps

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

$$5\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1\right)-\left(2\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$5\left(3-2\sqrt{3}+1\right)-\left(2\sqrt{3}\right)^{2}$$

Add 3 and 1 to get 4.

$$5\left(4-2\sqrt{3}\right)-\left(2\sqrt{3}\right)^{2}$$

Use the distributive property to multiply 5 by 4-2\sqrt{3}.

$$20-10\sqrt{3}-\left(2\sqrt{3}\right)^{2}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$20-10\sqrt{3}-2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$20-10\sqrt{3}-4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$20-10\sqrt{3}-4\times 3$$

Multiply 4 and 3 to get 12.

$$20-10\sqrt{3}-12$$

Subtract 12 from 20 to get 8.

$$8-10\sqrt{3}$$

Expand

$$8-10\sqrt{3}$$

Solution Steps

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

$$5\left(\left(\sqrt{3}\right)^{2}-2\sqrt{3}+1\right)-\left(2\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$5\left(3-2\sqrt{3}+1\right)-\left(2\sqrt{3}\right)^{2}$$

Add 3 and 1 to get 4.

$$5\left(4-2\sqrt{3}\right)-\left(2\sqrt{3}\right)^{2}$$

Use the distributive property to multiply 5 by 4-2\sqrt{3}.

$$20-10\sqrt{3}-\left(2\sqrt{3}\right)^{2}$$

Expand \left(2\sqrt{3}\right)^{2}.

$$20-10\sqrt{3}-2^{2}\left(\sqrt{3}\right)^{2}$$

Calculate 2 to the power of 2 and get 4.

$$20-10\sqrt{3}-4\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$20-10\sqrt{3}-4\times 3$$

Multiply 4 and 3 to get 12.

$$20-10\sqrt{3}-12$$

Subtract 12 from 20 to get 8.

$$8-10\sqrt{3}$$