Question
Answer
$$\dfrac{2-3\sqrt{18}}{\sqrt{18}+4}=21\sqrt{2}-31$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2-3\sqrt{18}}{\sqrt{18}+4}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2-3\sqrt{18}}{\sqrt{18}+4}\frac{\sqrt{18}-4}{\sqrt{18}-4} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{2}-8-54+36\sqrt{2}}{18-12\sqrt{2}+12\sqrt{2}-16} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{42\sqrt{2}-62}{2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{21\sqrt{2}-31}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}21\sqrt{2}-31\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{18}-4} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 2- 3 \sqrt{18}\right) } \cdot \left( \sqrt{18}-4\right) = \color{blue}{2} \cdot \sqrt{18}+\color{blue}{2} \cdot-4\color{blue}{- 3 \sqrt{18}} \cdot \sqrt{18}\color{blue}{- 3 \sqrt{18}} \cdot-4 = \\ = 6 \sqrt{2}-8-54 + 36 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{18} + 4\right) } \cdot \left( \sqrt{18}-4\right) = \color{blue}{ \sqrt{18}} \cdot \sqrt{18}+\color{blue}{ \sqrt{18}} \cdot-4+\color{blue}{4} \cdot \sqrt{18}+\color{blue}{4} \cdot-4 = \\ = 18- 12 \sqrt{2} + 12 \sqrt{2}-16 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 2. |
| ⑤ | Remove 1 from denominator. |
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