Question
Answer
$$\dfrac{6}{\sqrt{3}-7}=-\dfrac{6\sqrt{3}+42}{46}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{6}{\sqrt{3}-7}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{6}{\sqrt{3}-7}\frac{\sqrt{3}+7}{\sqrt{3}+7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{6\sqrt{3}+42}{3+7\sqrt{3}-7\sqrt{3}-49} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{6\sqrt{3}+42}{-46} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{6\sqrt{3}+42}{46}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{3} + 7} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ 6 } \cdot \left( \sqrt{3} + 7\right) = \color{blue}{6} \cdot \sqrt{3}+\color{blue}{6} \cdot7 = \\ = 6 \sqrt{3} + 42 $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{3}-7\right) } \cdot \left( \sqrt{3} + 7\right) = \color{blue}{ \sqrt{3}} \cdot \sqrt{3}+\color{blue}{ \sqrt{3}} \cdot7\color{blue}{-7} \cdot \sqrt{3}\color{blue}{-7} \cdot7 = \\ = 3 + 7 \sqrt{3}- 7 \sqrt{3}-49 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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