Solve quadx=frac{sqrt{7}+sqrt{3}}{sqrt{7}-sqrt{3}}

$$\quad x = \frac { \sqrt { 7 } + \sqrt { 3 } } { \sqrt { 7 } - \sqrt { 3 } }$$

Solve for x

$$x = \frac{\sqrt{21} + 5}{2} \approx 4.791287847$$

Steps for Solving Linear Equation

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

$$x=\frac{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}$$

Consider \left(\sqrt{7}-\sqrt{3}\right)\left(\sqrt{7}+\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}.

$$x=\frac{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}{\left(\sqrt{7}\right)^{2}-\left(\sqrt{3}\right)^{2}}$$

Square \sqrt{7}. Square \sqrt{3}.

$$x=\frac{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}{7-3}$$

Subtract 3 from 7 to get 4.

$$x=\frac{\left(\sqrt{7}+\sqrt{3}\right)\left(\sqrt{7}+\sqrt{3}\right)}{4}$$

Multiply \sqrt{7}+\sqrt{3} and \sqrt{7}+\sqrt{3} to get \left(\sqrt{7}+\sqrt{3}\right)^{2}.

$$x=\frac{\left(\sqrt{7}+\sqrt{3}\right)^{2}}{4}$$

Use binomial theorem \left(a+b\right)^{2}=a^{2}+2ab+b^{2} to expand \left(\sqrt{7}+\sqrt{3}\right)^{2}.

$$x=\frac{\left(\sqrt{7}\right)^{2}+2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{4}$$

The square of \sqrt{7} is 7.

$$x=\frac{7+2\sqrt{7}\sqrt{3}+\left(\sqrt{3}\right)^{2}}{4}$$

To multiply \sqrt{7} and \sqrt{3}, multiply the numbers under the square root.

$$x=\frac{7+2\sqrt{21}+\left(\sqrt{3}\right)^{2}}{4}$$

The square of \sqrt{3} is 3.

$$x=\frac{7+2\sqrt{21}+3}{4}$$

Add 7 and 3 to get 10.

$$x=\frac{10+2\sqrt{21}}{4}$$

Divide each term of 10+2\sqrt{21} by 4 to get \frac{5}{2}+\frac{1}{2}\sqrt{21}.

$$x=\frac{5}{2}+\frac{1}{2}\sqrt{21}$$

Assign x

$$x≔\frac{\sqrt{21}+5}{2}$$