Question
Answer
$$\dfrac{-20+3\sqrt{11}}{15+4\sqrt{11}}=\dfrac{-432+125\sqrt{11}}{49}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-20+3\sqrt{11}}{15+4\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-20+3\sqrt{11}}{15+4\sqrt{11}}\frac{15-4\sqrt{11}}{15-4\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-300+80\sqrt{11}+45\sqrt{11}-132}{225-60\sqrt{11}+60\sqrt{11}-176} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-432+125\sqrt{11}}{49}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 15- 4 \sqrt{11}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( -20 + 3 \sqrt{11}\right) } \cdot \left( 15- 4 \sqrt{11}\right) = \color{blue}{-20} \cdot15\color{blue}{-20} \cdot- 4 \sqrt{11}+\color{blue}{ 3 \sqrt{11}} \cdot15+\color{blue}{ 3 \sqrt{11}} \cdot- 4 \sqrt{11} = \\ = -300 + 80 \sqrt{11} + 45 \sqrt{11}-132 $$ Simplify denominator. $$ \color{blue}{ \left( 15 + 4 \sqrt{11}\right) } \cdot \left( 15- 4 \sqrt{11}\right) = \color{blue}{15} \cdot15+\color{blue}{15} \cdot- 4 \sqrt{11}+\color{blue}{ 4 \sqrt{11}} \cdot15+\color{blue}{ 4 \sqrt{11}} \cdot- 4 \sqrt{11} = \\ = 225- 60 \sqrt{11} + 60 \sqrt{11}-176 $$ |
| ③ | Simplify numerator and denominator |
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