Question
Answer
$$\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}=\dfrac{-119+31\sqrt{14}}{7}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{5-3\sqrt{14}}{7+2\sqrt{14}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{5-3\sqrt{14}}{7+2\sqrt{14}}\frac{7-2\sqrt{14}}{7-2\sqrt{14}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{35-10\sqrt{14}-21\sqrt{14}+84}{49-14\sqrt{14}+14\sqrt{14}-56} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{119-31\sqrt{14}}{-7} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{-119+31\sqrt{14}}{7}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 7- 2 \sqrt{14}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 5- 3 \sqrt{14}\right) } \cdot \left( 7- 2 \sqrt{14}\right) = \color{blue}{5} \cdot7+\color{blue}{5} \cdot- 2 \sqrt{14}\color{blue}{- 3 \sqrt{14}} \cdot7\color{blue}{- 3 \sqrt{14}} \cdot- 2 \sqrt{14} = \\ = 35- 10 \sqrt{14}- 21 \sqrt{14} + 84 $$ Simplify denominator. $$ \color{blue}{ \left( 7 + 2 \sqrt{14}\right) } \cdot \left( 7- 2 \sqrt{14}\right) = \color{blue}{7} \cdot7+\color{blue}{7} \cdot- 2 \sqrt{14}+\color{blue}{ 2 \sqrt{14}} \cdot7+\color{blue}{ 2 \sqrt{14}} \cdot- 2 \sqrt{14} = \\ = 49- 14 \sqrt{14} + 14 \sqrt{14}-56 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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