Question
Answer
$$\dfrac{7}{2\sqrt{2}+1}=2\sqrt{2}-1$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{7}{2\sqrt{2}+1}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{7}{2\sqrt{2}+1}\frac{2\sqrt{2}-1}{2\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{14\sqrt{2}-7}{8-2\sqrt{2}+2\sqrt{2}-1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{14\sqrt{2}-7}{7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{2\sqrt{2}-1}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}2\sqrt{2}-1\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{2}-1} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 7 } \cdot \left( 2 \sqrt{2}-1\right) = \color{blue}{7} \cdot 2 \sqrt{2}+\color{blue}{7} \cdot-1 = \\ = 14 \sqrt{2}-7 $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{2} + 1\right) } \cdot \left( 2 \sqrt{2}-1\right) = \color{blue}{ 2 \sqrt{2}} \cdot 2 \sqrt{2}+\color{blue}{ 2 \sqrt{2}} \cdot-1+\color{blue}{1} \cdot 2 \sqrt{2}+\color{blue}{1} \cdot-1 = \\ = 8- 2 \sqrt{2} + 2 \sqrt{2}-1 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 7. |
| ⑤ | Remove 1 from denominator. |
This page was created using
Rationalize Denominator Calculator