Question
Answer
$$\dfrac{\sqrt{3}}{1-4\sqrt{2}}=-\dfrac{\sqrt{3}+4\sqrt{6}}{31}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}}{1-4\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}}{1-4\sqrt{2}}\frac{1+4\sqrt{2}}{1+4\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{\sqrt{3}+4\sqrt{6}}{1+4\sqrt{2}-4\sqrt{2}-32} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{\sqrt{3}+4\sqrt{6}}{-31} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{\sqrt{3}+4\sqrt{6}}{31}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 1 + 4 \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{3} } \cdot \left( 1 + 4 \sqrt{2}\right) = \color{blue}{ \sqrt{3}} \cdot1+\color{blue}{ \sqrt{3}} \cdot 4 \sqrt{2} = \\ = \sqrt{3} + 4 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 1- 4 \sqrt{2}\right) } \cdot \left( 1 + 4 \sqrt{2}\right) = \color{blue}{1} \cdot1+\color{blue}{1} \cdot 4 \sqrt{2}\color{blue}{- 4 \sqrt{2}} \cdot1\color{blue}{- 4 \sqrt{2}} \cdot 4 \sqrt{2} = \\ = 1 + 4 \sqrt{2}- 4 \sqrt{2}-32 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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