Solve {l}{(12+3sqrt{3}}{-sqrt{48})^2}

$$\left. \begin{array} { l } { ( 12 + 3 \sqrt { 3 } } \\ { - \sqrt { 48 } ) ^ { 2 } } \end{array} \right.$$

Evaluate

$$147-24\sqrt{3}\approx 105.430780618$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-\sqrt{3}\right)^{2}.

$$144-24\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$144-24\sqrt{3}+3$$

Add 144 and 3 to get 147.

$$147-24\sqrt{3}$$

Expand

$$147-24\sqrt{3}$$

Solution Steps

Factor 48=4^{2}\times 3. Rewrite the square root of the product \sqrt{4^{2}\times 3} as the product of square roots \sqrt{4^{2}}\sqrt{3}. Take the square root of 4^{2}.

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

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

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

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(12-\sqrt{3}\right)^{2}.

$$144-24\sqrt{3}+\left(\sqrt{3}\right)^{2}$$

The square of \sqrt{3} is 3.

$$144-24\sqrt{3}+3$$

Add 144 and 3 to get 147.

$$147-24\sqrt{3}$$