Solve 14:(sqrt{3}-2)^2+sqrt{12}+6sqrt{1/3}

$$14 : ( \sqrt { 3 } - 2 ) ^ { 2 } + \sqrt { 12 } + 6 \sqrt { \frac { 1 } { 3 } }$$

Evaluate

$$60\sqrt{3}+98\approx 201.923048454$$

Solution Steps

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

$$\frac{14}{\left(\sqrt{3}\right)^{2}-4\sqrt{3}+4}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

The square of \sqrt{3} is 3.

$$\frac{14}{3-4\sqrt{3}+4}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Add 3 and 4 to get 7.

$$\frac{14}{7-4\sqrt{3}}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Rationalize the denominator of \frac{14}{7-4\sqrt{3}} by multiplying numerator and denominator by 7+4\sqrt{3}.

$$\frac{14\left(7+4\sqrt{3}\right)}{\left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right)}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Consider \left(7-4\sqrt{3}\right)\left(7+4\sqrt{3}\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{14\left(7+4\sqrt{3}\right)}{7^{2}-\left(-4\sqrt{3}\right)^{2}}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Calculate 7 to the power of 2 and get 49.

$$\frac{14\left(7+4\sqrt{3}\right)}{49-\left(-4\sqrt{3}\right)^{2}}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

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

$$\frac{14\left(7+4\sqrt{3}\right)}{49-\left(-4\right)^{2}\left(\sqrt{3}\right)^{2}}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Calculate -4 to the power of 2 and get 16.

$$\frac{14\left(7+4\sqrt{3}\right)}{49-16\left(\sqrt{3}\right)^{2}}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

The square of \sqrt{3} is 3.

$$\frac{14\left(7+4\sqrt{3}\right)}{49-16\times 3}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Multiply 16 and 3 to get 48.

$$\frac{14\left(7+4\sqrt{3}\right)}{49-48}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Subtract 48 from 49 to get 1.

$$\frac{14\left(7+4\sqrt{3}\right)}{1}+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

Anything divided by one gives itself.

$$14\left(7+4\sqrt{3}\right)+\sqrt{12}+6\sqrt{\frac{1}{3}}$$

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}.

$$14\left(7+4\sqrt{3}\right)+2\sqrt{3}+6\sqrt{\frac{1}{3}}$$

Rewrite the square root of the division \sqrt{\frac{1}{3}} as the division of square roots \frac{\sqrt{1}}{\sqrt{3}}.

$$14\left(7+4\sqrt{3}\right)+2\sqrt{3}+6\times \frac{\sqrt{1}}{\sqrt{3}}$$

Calculate the square root of 1 and get 1.

$$14\left(7+4\sqrt{3}\right)+2\sqrt{3}+6\times \frac{1}{\sqrt{3}}$$

Rationalize the denominator of \frac{1}{\sqrt{3}} by multiplying numerator and denominator by \sqrt{3}.

$$14\left(7+4\sqrt{3}\right)+2\sqrt{3}+6\times \frac{\sqrt{3}}{\left(\sqrt{3}\right)^{2}}$$

The square of \sqrt{3} is 3.

$$14\left(7+4\sqrt{3}\right)+2\sqrt{3}+6\times \frac{\sqrt{3}}{3}$$

Cancel out 3, the greatest common factor in 6 and 3.

$$14\left(7+4\sqrt{3}\right)+2\sqrt{3}+2\sqrt{3}$$

Combine 2\sqrt{3} and 2\sqrt{3} to get 4\sqrt{3}.

$$14\left(7+4\sqrt{3}\right)+4\sqrt{3}$$

Use the distributive property to multiply 14 by 7+4\sqrt{3}.

$$98+56\sqrt{3}+4\sqrt{3}$$

Combine 56\sqrt{3} and 4\sqrt{3} to get 60\sqrt{3}.

$$98+60\sqrt{3}$$