Solve sqrt{48}:sqrt{3}-2(sqrt{2})^2+2(sqrt{3})^2:sqrt{2^2}+4

$$\sqrt { 48 } : \sqrt { 3 } - 2 ( \sqrt { 2 } ) ^ { 2 } + 2 ( \sqrt { 3 } ) ^ { 2 } : \sqrt { 2 ^ { 2 } } + 4$$

Evaluate

$$7$$

Solution Steps

Rewrite the division of square roots \frac{\sqrt{48}}{\sqrt{3}} as the square root of the division \sqrt{\frac{48}{3}} and perform the division.

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

Calculate the square root of 16 and get 4.

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

The square of \sqrt{2} is 2.

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

Multiply 2 and 2 to get 4.

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

Subtract 4 from 4 to get 0.

$$0+\frac{2\left(\sqrt{3}\right)^{2}}{\sqrt{2^{2}}}+4$$

The square of \sqrt{3} is 3.

$$0+\frac{2\times 3}{\sqrt{2^{2}}}+4$$

Multiply 2 and 3 to get 6.

$$0+\frac{6}{\sqrt{2^{2}}}+4$$

Calculate 2 to the power of 2 and get 4.

$$0+\frac{6}{\sqrt{4}}+4$$

Calculate the square root of 4 and get 2.

$$0+\frac{6}{2}+4$$

Divide 6 by 2 to get 3.

$$0+3+4$$

Add 0 and 3 to get 3.

$$3+4$$

Add 3 and 4 to get 7.

$$7$$

Factor

$$7$$