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

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

Evaluate

$$6$$

Solution Steps

Add 1 and 2 to get 3.

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

Calculate 3 to the power of 2 and get 9.

$$\sqrt{9+\left(-\sqrt{3}-2\sqrt{3}\right)^{2}}$$

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

$$\sqrt{9+\left(-\sqrt{3}\right)^{2}-4\left(-\sqrt{3}\right)\sqrt{3}+4\left(\sqrt{3}\right)^{2}}$$

Calculate -\sqrt{3} to the power of 2 and get \left(\sqrt{3}\right)^{2}.

$$\sqrt{9+\left(\sqrt{3}\right)^{2}-4\left(-\sqrt{3}\right)\sqrt{3}+4\left(\sqrt{3}\right)^{2}}$$

Multiply -4 and -1 to get 4.

$$\sqrt{9+\left(\sqrt{3}\right)^{2}+4\sqrt{3}\sqrt{3}+4\left(\sqrt{3}\right)^{2}}$$

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

$$\sqrt{9+\left(\sqrt{3}\right)^{2}+4\times 3+4\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$\sqrt{9+\left(\sqrt{3}\right)^{2}+4\times 3+4\times 3}$$

Multiply 4 and 3 to get 12.

$$\sqrt{9+\left(\sqrt{3}\right)^{2}+4\times 3+12}$$

Multiply 4 and 3 to get 12.

$$\sqrt{9+\left(\sqrt{3}\right)^{2}+12+12}$$

Add 9 and 12 to get 21.

$$\sqrt{21+\left(\sqrt{3}\right)^{2}+12}$$

The square of \sqrt{3} is 3.

$$\sqrt{21+3+12}$$

Add 21 and 3 to get 24.

$$\sqrt{24+12}$$

Add 24 and 12 to get 36.

$$\sqrt{36}$$

Calculate the square root of 36 and get 6.

$$6$$

Factor

$$2\times 3$$