Question
Answer
$$\dfrac{\sqrt{3}-\sqrt{5}}{2\sqrt{7}+3\sqrt{3}}=2\sqrt{21}-9-2\sqrt{35}+3\sqrt{15}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}-\sqrt{5}}{2\sqrt{7}+3\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}-\sqrt{5}}{2\sqrt{7}+3\sqrt{3}}\frac{2\sqrt{7}-3\sqrt{3}}{2\sqrt{7}-3\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{21}-9-2\sqrt{35}+3\sqrt{15}}{28-6\sqrt{21}+6\sqrt{21}-27} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{2\sqrt{21}-9-2\sqrt{35}+3\sqrt{15}}{1} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}2\sqrt{21}-9-2\sqrt{35}+3\sqrt{15}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2 \sqrt{7}- 3 \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3}- \sqrt{5}\right) } \cdot \left( 2 \sqrt{7}- 3 \sqrt{3}\right) = \color{blue}{ \sqrt{3}} \cdot 2 \sqrt{7}+\color{blue}{ \sqrt{3}} \cdot- 3 \sqrt{3}\color{blue}{- \sqrt{5}} \cdot 2 \sqrt{7}\color{blue}{- \sqrt{5}} \cdot- 3 \sqrt{3} = \\ = 2 \sqrt{21}-9- 2 \sqrt{35} + 3 \sqrt{15} $$ Simplify denominator. $$ \color{blue}{ \left( 2 \sqrt{7} + 3 \sqrt{3}\right) } \cdot \left( 2 \sqrt{7}- 3 \sqrt{3}\right) = \color{blue}{ 2 \sqrt{7}} \cdot 2 \sqrt{7}+\color{blue}{ 2 \sqrt{7}} \cdot- 3 \sqrt{3}+\color{blue}{ 3 \sqrt{3}} \cdot 2 \sqrt{7}+\color{blue}{ 3 \sqrt{3}} \cdot- 3 \sqrt{3} = \\ = 28- 6 \sqrt{21} + 6 \sqrt{21}-27 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Remove 1 from denominator. |
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