Question
Answer
$$\dfrac{-\sqrt{35}}{\sqrt{5}+\sqrt{35}}=\dfrac{\sqrt{7}-7}{6}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-\sqrt{35}}{\sqrt{5}+\sqrt{35}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-\sqrt{35}}{\sqrt{5}+\sqrt{35}}\frac{\sqrt{5}-\sqrt{35}}{\sqrt{5}-\sqrt{35}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-5\sqrt{7}+35}{5-5\sqrt{7}+5\sqrt{7}-35} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-5\sqrt{7}+35}{-30} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-\sqrt{7}+7}{-6} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{\sqrt{7}-7}{6}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{5}- \sqrt{35}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ - \sqrt{35} } \cdot \left( \sqrt{5}- \sqrt{35}\right) = \color{blue}{- \sqrt{35}} \cdot \sqrt{5}\color{blue}{- \sqrt{35}} \cdot- \sqrt{35} = \\ = - 5 \sqrt{7} + 35 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{5} + \sqrt{35}\right) } \cdot \left( \sqrt{5}- \sqrt{35}\right) = \color{blue}{ \sqrt{5}} \cdot \sqrt{5}+\color{blue}{ \sqrt{5}} \cdot- \sqrt{35}+\color{blue}{ \sqrt{35}} \cdot \sqrt{5}+\color{blue}{ \sqrt{35}} \cdot- \sqrt{35} = \\ = 5- 5 \sqrt{7} + 5 \sqrt{7}-35 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Divide both numerator and denominator by 5. |
| ⑤ | Multiply both numerator and denominator by -1. |
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