Question
Answer
$$\dfrac{\sqrt{10}}{7+\sqrt{15}}=\dfrac{7\sqrt{10}-5\sqrt{6}}{34}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{10}}{7+\sqrt{15}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{10}}{7+\sqrt{15}}\frac{7-\sqrt{15}}{7-\sqrt{15}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7\sqrt{10}-5\sqrt{6}}{49-7\sqrt{15}+7\sqrt{15}-15} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7\sqrt{10}-5\sqrt{6}}{34}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- \sqrt{15}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{10} } \cdot \left( 7- \sqrt{15}\right) = \color{blue}{ \sqrt{10}} \cdot7+\color{blue}{ \sqrt{10}} \cdot- \sqrt{15} = \\ = 7 \sqrt{10}- 5 \sqrt{6} $$ Simplify denominator. $$ \color{blue}{ \left( 7 + \sqrt{15}\right) } \cdot \left( 7- \sqrt{15}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- \sqrt{15}+\color{blue}{ \sqrt{15}} \cdot7+\color{blue}{ \sqrt{15}} \cdot- \sqrt{15} = \\ = 49- 7 \sqrt{15} + 7 \sqrt{15}-15 $$ |
| ③ | Simplify numerator and denominator |
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