Question
Answer
$$\dfrac{\sqrt{3}}{\sqrt{5}-\sqrt{14}}=-\dfrac{\sqrt{15}+\sqrt{42}}{9}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{\sqrt{5}-\sqrt{14}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{\sqrt{5}-\sqrt{14}}\frac{\sqrt{5}+\sqrt{14}}{\sqrt{5}+\sqrt{14}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{15}+\sqrt{42}}{5+\sqrt{70}-\sqrt{70}-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{15}+\sqrt{42}}{-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{\sqrt{15}+\sqrt{42}}{9}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5} + \sqrt{14}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( \sqrt{5} + \sqrt{14}\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{5}+\color{blue}{ \sqrt{3}} \cdot \sqrt{14} = \\ = \sqrt{15} + \sqrt{42} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5}- \sqrt{14}\right) } \cdot \left( \sqrt{5} + \sqrt{14}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot \sqrt{14}\color{blue}{- \sqrt{14}} \cdot \sqrt{5}\color{blue}{- \sqrt{14}} \cdot \sqrt{14} = \\ = 5 + \sqrt{70}- \sqrt{70}-14 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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