Answer
$$\frac{\sqrt{7}}{\sqrt{7}-3}=-\frac{7+3\sqrt{7}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{7}}{\sqrt{7}-3}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{7}}{\sqrt{7}-3}\frac{\sqrt{7}+3}{\sqrt{7}+3} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{7+3\sqrt{7}}{7+3\sqrt{7}-3\sqrt{7}-9} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{7+3\sqrt{7}}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}-\frac{7+3\sqrt{7}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{7} + 3} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{7} } \cdot \left( \sqrt{7} + 3\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot3 = \\ = 7 + 3 \sqrt{7} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{7}-3\right) } \cdot \left( \sqrt{7} + 3\right) = \color{blue}{ \sqrt{7}} \cdot \sqrt{7}+\color{blue}{ \sqrt{7}} \cdot3\color{blue}{-3} \cdot \sqrt{7}\color{blue}{-3} \cdot3 = \\ = 7 + 3 \sqrt{7}- 3 \sqrt{7}-9 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Place a negative sign in front of a fraction. |
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