Question
Answer
$$\dfrac{2\sqrt{2}}{\sqrt{3}-8}=-\dfrac{2\sqrt{6}+16\sqrt{2}}{61}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{2}}{\sqrt{3}-8}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{2}}{\sqrt{3}-8}\frac{\sqrt{3}+8}{\sqrt{3}+8} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{6}+16\sqrt{2}}{3+8\sqrt{3}-8\sqrt{3}-64} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{6}+16\sqrt{2}}{-61} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{2\sqrt{6}+16\sqrt{2}}{61}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + 8} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 2 \sqrt{2} } \cdot \left( \sqrt{3} + 8\right) = \color{blue}{ 2 \sqrt{2}} \cdot \sqrt{3}+\color{blue}{ 2 \sqrt{2}} \cdot8 = \\ = 2 \sqrt{6} + 16 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}-8\right) } \cdot \left( \sqrt{3} + 8\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot8\color{blue}{-8} \cdot \sqrt{3}\color{blue}{-8} \cdot8 = \\ = 3 + 8 \sqrt{3}- 8 \sqrt{3}-64 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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