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