Question
Answer
$$\dfrac{4+\sqrt{3}}{6-\sqrt{7}}=\dfrac{24+4\sqrt{7}+6\sqrt{3}+\sqrt{21}}{29}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{4+\sqrt{3}}{6-\sqrt{7}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{4+\sqrt{3}}{6-\sqrt{7}}\frac{6+\sqrt{7}}{6+\sqrt{7}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{24+4\sqrt{7}+6\sqrt{3}+\sqrt{21}}{36+6\sqrt{7}-6\sqrt{7}-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{24+4\sqrt{7}+6\sqrt{3}+\sqrt{21}}{29}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 6 + \sqrt{7}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 4 + \sqrt{3}\right) } \cdot \left( 6 + \sqrt{7}\right) = \color{blue}{4} \cdot6+\color{blue}{4} \cdot \sqrt{7}+\color{blue}{ \sqrt{3}} \cdot6+\color{blue}{ \sqrt{3}} \cdot \sqrt{7} = \\ = 24 + 4 \sqrt{7} + 6 \sqrt{3} + \sqrt{21} $$ Simplify denominator. $$ \color{blue}{ \left( 6- \sqrt{7}\right) } \cdot \left( 6 + \sqrt{7}\right) = \color{blue}{6} \cdot6+\color{blue}{6} \cdot \sqrt{7}\color{blue}{- \sqrt{7}} \cdot6\color{blue}{- \sqrt{7}} \cdot \sqrt{7} = \\ = 36 + 6 \sqrt{7}- 6 \sqrt{7}-7 $$ |
| ③ | Simplify numerator and denominator |
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