Question
Answer
$$\dfrac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}=6+\sqrt{35}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}-\sqrt{5}}\frac{\sqrt{7}+\sqrt{5}}{\sqrt{7}+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7+\sqrt{35}+\sqrt{35}+5}{7+\sqrt{35}-\sqrt{35}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12+2\sqrt{35}}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{6+\sqrt{35}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}6+\sqrt{35}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7} + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{7} + \sqrt{5}\right) } \cdot \left( \sqrt{7} + \sqrt{5}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot \sqrt{7}+\color{blue}{ \sqrt{5}} \cdot \sqrt{5} = \\ = 7 + \sqrt{35} + \sqrt{35} + 5 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7}- \sqrt{5}\right) } \cdot \left( \sqrt{7} + \sqrt{5}\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot \sqrt{7}\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 7 + \sqrt{35}- \sqrt{35}-5 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Remove 1 from denominator. |
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