Solve (sqrt{3}+sqrt{27})^2

$$( \sqrt { 3 } + \sqrt { 27 } ) ^ { 2 }$$

Evaluate

$$48$$

Solution Steps

Factor 27=3^{2}\times 3. Rewrite the square root of the product \sqrt{3^{2}\times 3} as the product of square roots \sqrt{3^{2}}\sqrt{3}. Take the square root of 3^{2}.

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

Combine \sqrt{3} and 3\sqrt{3} to get 4\sqrt{3}.

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

Expand \left(4\sqrt{3}\right)^{2}.

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

Calculate 4 to the power of 2 and get 16.

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

The square of \sqrt{3} is 3.

$$16\times 3$$

Multiply 16 and 3 to get 48.

$$48$$

Factor

$$2^{4}\times 3$$