Question
Answer
$$\dfrac{5}{\sqrt{35}+\sqrt{14}}=\dfrac{5\sqrt{35}-5\sqrt{14}}{21}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5}{\sqrt{35}+\sqrt{14}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5}{\sqrt{35}+\sqrt{14}}\frac{\sqrt{35}-\sqrt{14}}{\sqrt{35}-\sqrt{14}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{5\sqrt{35}-5\sqrt{14}}{35-7\sqrt{10}+7\sqrt{10}-14} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{5\sqrt{35}-5\sqrt{14}}{21}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{35}- \sqrt{14}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 5 } \cdot \left( \sqrt{35}- \sqrt{14}\right) = \color{blue}{5} \cdot \sqrt{35}+\color{blue}{5} \cdot- \sqrt{14} = \\ = 5 \sqrt{35}- 5 \sqrt{14} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{35} + \sqrt{14}\right) } \cdot \left( \sqrt{35}- \sqrt{14}\right) = \color{blue}{ \sqrt{35}} \cdot \sqrt{35}+\color{blue}{ \sqrt{35}} \cdot- \sqrt{14}+\color{blue}{ \sqrt{14}} \cdot \sqrt{35}+\color{blue}{ \sqrt{14}} \cdot- \sqrt{14} = \\ = 35- 7 \sqrt{10} + 7 \sqrt{10}-14 $$ |
| ③ | Simplify numerator and denominator |
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