Question
Answer
$$\dfrac{-3}{\sqrt{35}+\sqrt{3}}=\dfrac{-3\sqrt{35}+3\sqrt{3}}{32}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-3}{\sqrt{35}+\sqrt{3}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-3}{\sqrt{35}+\sqrt{3}}\frac{\sqrt{35}-\sqrt{3}}{\sqrt{35}-\sqrt{3}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-3\sqrt{35}+3\sqrt{3}}{35-\sqrt{105}+\sqrt{105}-3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-3\sqrt{35}+3\sqrt{3}}{32}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{35}- \sqrt{3}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -3 } \cdot \left( \sqrt{35}- \sqrt{3}\right) = \color{blue}{-3} \cdot \sqrt{35}\color{blue}{-3} \cdot- \sqrt{3} = \\ = - 3 \sqrt{35} + 3 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{35} + \sqrt{3}\right) } \cdot \left( \sqrt{35}- \sqrt{3}\right) = \color{blue}{ \sqrt{35}} \cdot \sqrt{35}+\color{blue}{ \sqrt{35}} \cdot- \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot \sqrt{35}+\color{blue}{ \sqrt{3}} \cdot- \sqrt{3} = \\ = 35- \sqrt{105} + \sqrt{105}-3 $$ |
| ③ | Simplify numerator and denominator |
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