Question
Answer
$$\dfrac{\sqrt{3}}{\sqrt{14}-\sqrt{5}}=\dfrac{\sqrt{42}+\sqrt{15}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{\sqrt{14}-\sqrt{5}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{\sqrt{14}-\sqrt{5}}\frac{\sqrt{14}+\sqrt{5}}{\sqrt{14}+\sqrt{5}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{42}+\sqrt{15}}{14+\sqrt{70}-\sqrt{70}-5} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{42}+\sqrt{15}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{14} + \sqrt{5}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( \sqrt{14} + \sqrt{5}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{14}+\color{blue}{ \sqrt{3}} \cdot \sqrt{5} = \\ = \sqrt{42} + \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{14}- \sqrt{5}\right) } \cdot \left( \sqrt{14} + \sqrt{5}\right) = \color{blue}{ \sqrt{14}} \cdot \sqrt{14}+\color{blue}{ \sqrt{14}} \cdot \sqrt{5}\color{blue}{- \sqrt{5}} \cdot \sqrt{14}\color{blue}{- \sqrt{5}} \cdot \sqrt{5} = \\ = 14 + \sqrt{70}- \sqrt{70}-5 $$ |
| ③ | Simplify numerator and denominator |
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