Solve frac{2sqrt{3}-5}{2sqrt{3}+5}

$$\frac { 2 \sqrt { 3 } - 5 } { 2 \sqrt { 3 } + 5 }$$

Evaluate

$$\frac{20\sqrt{3}-37}{13}\approx -0.181460296$$

Solution Steps

Rationalize the denominator of \frac{2\sqrt{3}-5}{2\sqrt{3}+5} by multiplying numerator and denominator by 2\sqrt{3}-5.

$$\frac{\left(2\sqrt{3}-5\right)\left(2\sqrt{3}-5\right)}{\left(2\sqrt{3}+5\right)\left(2\sqrt{3}-5\right)}$$

Consider \left(2\sqrt{3}+5\right)\left(2\sqrt{3}-5\right). Multiplication can be transformed into difference of squares using the rule: \left(a-b\right)\left(a+b\right)=a^{2}-b^{2}.

$$\frac{\left(2\sqrt{3}-5\right)\left(2\sqrt{3}-5\right)}{\left(2\sqrt{3}\right)^{2}-5^{2}}$$

Multiply 2\sqrt{3}-5 and 2\sqrt{3}-5 to get \left(2\sqrt{3}-5\right)^{2}.

$$\frac{\left(2\sqrt{3}-5\right)^{2}}{\left(2\sqrt{3}\right)^{2}-5^{2}}$$

Use binomial theorem \left(a-b\right)^{2}=a^{2}-2ab+b^{2} to expand \left(2\sqrt{3}-5\right)^{2}.

$$\frac{4\left(\sqrt{3}\right)^{2}-20\sqrt{3}+25}{\left(2\sqrt{3}\right)^{2}-5^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{4\times 3-20\sqrt{3}+25}{\left(2\sqrt{3}\right)^{2}-5^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{12-20\sqrt{3}+25}{\left(2\sqrt{3}\right)^{2}-5^{2}}$$

Add 12 and 25 to get 37.

$$\frac{37-20\sqrt{3}}{\left(2\sqrt{3}\right)^{2}-5^{2}}$$

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

$$\frac{37-20\sqrt{3}}{2^{2}\left(\sqrt{3}\right)^{2}-5^{2}}$$

Calculate 2 to the power of 2 and get 4.

$$\frac{37-20\sqrt{3}}{4\left(\sqrt{3}\right)^{2}-5^{2}}$$

The square of \sqrt{3} is 3.

$$\frac{37-20\sqrt{3}}{4\times 3-5^{2}}$$

Multiply 4 and 3 to get 12.

$$\frac{37-20\sqrt{3}}{12-5^{2}}$$

Calculate 5 to the power of 2 and get 25.

$$\frac{37-20\sqrt{3}}{12-25}$$

Subtract 25 from 12 to get -13.

$$\frac{37-20\sqrt{3}}{-13}$$

Multiply both numerator and denominator by -1.

$$\frac{-37+20\sqrt{3}}{13}$$