Solve {left(2(sqrt{12}-3)right)}^2divsqrt{12}

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

Evaluate

$$14\sqrt{3}-24\approx 0.248711306$$

Solution Steps

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

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

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

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

Calculate 2 to the power of 2 and get 4.

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

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

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

The square of \sqrt{3} is 3.

$$\frac{4\left(4\times 3-12\sqrt{3}+9\right)}{\sqrt{12}}$$

Multiply 4 and 3 to get 12.

$$\frac{4\left(12-12\sqrt{3}+9\right)}{\sqrt{12}}$$

Add 12 and 9 to get 21.

$$\frac{4\left(21-12\sqrt{3}\right)}{\sqrt{12}}$$

Factor 12=2^{2}\times 3. Rewrite the square root of the product \sqrt{2^{2}\times 3} as the product of square roots \sqrt{2^{2}}\sqrt{3}. Take the square root of 2^{2}.

$$\frac{4\left(21-12\sqrt{3}\right)}{2\sqrt{3}}$$

Cancel out 2 in both numerator and denominator.

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

Rationalize the denominator of \frac{2\left(-12\sqrt{3}+21\right)}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

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

The square of \sqrt{3} is 3.

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

Use the distributive property to multiply 2 by -12\sqrt{3}+21.

$$\frac{\left(-24\sqrt{3}+42\right)\sqrt{3}}{3}$$

Use the distributive property to multiply -24\sqrt{3}+42 by \sqrt{3}.

$$\frac{-24\left(\sqrt{3}\right)^{2}+42\sqrt{3}}{3}$$

The square of \sqrt{3} is 3.

$$\frac{-24\times 3+42\sqrt{3}}{3}$$

Multiply -24 and 3 to get -72.

$$\frac{-72+42\sqrt{3}}{3}$$