Solve q=frac{2-sqrt{5}}{2+sqrt{5}}

$$q = \frac { 2 - \sqrt { 5 } } { 2 + \sqrt { 5 } }$$

Solve for q

$$q=4\sqrt{5}-9\approx -0.05572809$$

Steps for Solving Linear Equation

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

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

Consider \left(2+\sqrt{5}\right)\left(2-\sqrt{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}.

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

Square 2. Square \sqrt{5}.

$$q=\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{4-5}$$

Subtract 5 from 4 to get -1.

$$q=\frac{\left(2-\sqrt{5}\right)\left(2-\sqrt{5}\right)}{-1}$$

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

$$q=\frac{\left(2-\sqrt{5}\right)^{2}}{-1}$$

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

$$q=\frac{4-4\sqrt{5}+\left(\sqrt{5}\right)^{2}}{-1}$$

The square of \sqrt{5} is 5.

$$q=\frac{4-4\sqrt{5}+5}{-1}$$

Add 4 and 5 to get 9.

$$q=\frac{9-4\sqrt{5}}{-1}$$

Anything divided by -1 gives its opposite. To find the opposite of 9-4\sqrt{5}, find the opposite of each term.

$$q=-9-\left(-4\sqrt{5}\right)$$

The opposite of -4\sqrt{5} is 4\sqrt{5}.

$$q=-9+4\sqrt{5}$$

Assign q

$$q≔4\sqrt{5}-9$$