Question
Answer
$$\dfrac{15-3\sqrt{11}}{4+\sqrt{11}}=\dfrac{93-27\sqrt{11}}{5}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{15-3\sqrt{11}}{4+\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{15-3\sqrt{11}}{4+\sqrt{11}}\frac{4-\sqrt{11}}{4-\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{60-15\sqrt{11}-12\sqrt{11}+33}{16-4\sqrt{11}+4\sqrt{11}-11} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{93-27\sqrt{11}}{5}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 4- \sqrt{11}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( 15- 3 \sqrt{11}\right) } \cdot \left( 4- \sqrt{11}\right) = \color{blue}{15} \cdot4+\color{blue}{15} \cdot- \sqrt{11}\color{blue}{- 3 \sqrt{11}} \cdot4\color{blue}{- 3 \sqrt{11}} \cdot- \sqrt{11} = \\ = 60- 15 \sqrt{11}- 12 \sqrt{11} + 33 $$ Simplify denominator. $$ \color{blue}{ \left( 4 + \sqrt{11}\right) } \cdot \left( 4- \sqrt{11}\right) = \color{blue}{4} \cdot4+\color{blue}{4} \cdot- \sqrt{11}+\color{blue}{ \sqrt{11}} \cdot4+\color{blue}{ \sqrt{11}} \cdot- \sqrt{11} = \\ = 16- 4 \sqrt{11} + 4 \sqrt{11}-11 $$ |
| ③ | Simplify numerator and denominator |
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