Solve {left(1/sqrt{3-2}right)}^2

$${ \left( \frac{ 1 }{ \sqrt{ 3 } -2 } \right) }^{ 2 }$$

Evaluate

$$4\sqrt{3}+7\approx 13.92820323$$

Solution Steps

Rationalize the denominator of \frac{1}{\sqrt{3}-2} by multiplying numerator and denominator by \sqrt{3}+2.

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

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

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

Square \sqrt{3}. Square 2.

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

Subtract 4 from 3 to get -1.

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

Anything divided by -1 gives its opposite. To find the opposite of \sqrt{3}+2, find the opposite of each term.

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

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

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

The square of \sqrt{3} is 3.

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

Add 3 and 4 to get 7.

$$7+4\sqrt{3}$$