Question
Answer
$$\dfrac{-15+4\sqrt{11}}{20+3\sqrt{11}}=\dfrac{-432+125\sqrt{11}}{301}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-15+4\sqrt{11}}{20+3\sqrt{11}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-15+4\sqrt{11}}{20+3\sqrt{11}}\frac{20-3\sqrt{11}}{20-3\sqrt{11}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-300+45\sqrt{11}+80\sqrt{11}-132}{400-60\sqrt{11}+60\sqrt{11}-99} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-432+125\sqrt{11}}{301}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 20- 3 \sqrt{11}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( -15 + 4 \sqrt{11}\right) } \cdot \left( 20- 3 \sqrt{11}\right) = \color{blue}{-15} \cdot20\color{blue}{-15} \cdot- 3 \sqrt{11}+\color{blue}{ 4 \sqrt{11}} \cdot20+\color{blue}{ 4 \sqrt{11}} \cdot- 3 \sqrt{11} = \\ = -300 + 45 \sqrt{11} + 80 \sqrt{11}-132 $$ Simplify denominator. $$ \color{blue}{ \left( 20 + 3 \sqrt{11}\right) } \cdot \left( 20- 3 \sqrt{11}\right) = \color{blue}{20} \cdot20+\color{blue}{20} \cdot- 3 \sqrt{11}+\color{blue}{ 3 \sqrt{11}} \cdot20+\color{blue}{ 3 \sqrt{11}} \cdot- 3 \sqrt{11} = \\ = 400- 60 \sqrt{11} + 60 \sqrt{11}-99 $$ |
| ③ | Simplify numerator and denominator |
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