Solve (frac{sqrt{3}+1}{sqrt{3}-1})^2

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

Evaluate

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

Solution Steps

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

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Multiply \sqrt{3}+1 and \sqrt{3}+1 to get \left(\sqrt{3}+1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Divide each term of 4+2\sqrt{3} by 2 to get 2+\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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

Expand

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

Solution Steps

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

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

Consider \left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

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

Square \sqrt{3}. Square 1.

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

Subtract 1 from 3 to get 2.

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

Multiply \sqrt{3}+1 and \sqrt{3}+1 to get \left(\sqrt{3}+1\right)^{2}.

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

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{3}+1\right)^{2}.

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

The square of \sqrt{3} is 3.

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

Add 3 and 1 to get 4.

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

Divide each term of 4+2\sqrt{3} by 2 to get 2+\sqrt{3}.

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

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

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

The square of \sqrt{3} is 3.

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

Add 4 and 3 to get 7.

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