Question
Answer
$$\dfrac{\sqrt{3}+1}{3+\sqrt{48}}=\dfrac{\sqrt{3}+9}{39}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{3}+1}{3+\sqrt{48}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{3}+1}{3+\sqrt{48}}\frac{3-\sqrt{48}}{3-\sqrt{48}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{3\sqrt{3}-12+3-4\sqrt{3}}{9-12\sqrt{3}+12\sqrt{3}-48} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-\sqrt{3}-9}{-39} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{3}+9}{39}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 3- \sqrt{48}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \left( \sqrt{3} + 1\right) } \cdot \left( 3- \sqrt{48}\right) = \color{blue}{ \sqrt{3}} \cdot3+\color{blue}{ \sqrt{3}} \cdot- \sqrt{48}+\color{blue}{1} \cdot3+\color{blue}{1} \cdot- \sqrt{48} = \\ = 3 \sqrt{3}-12 + 3- 4 \sqrt{3} $$ Simplify denominator. $$ \color{blue}{ \left( 3 + \sqrt{48}\right) } \cdot \left( 3- \sqrt{48}\right) = \color{blue}{3} \cdot3+\color{blue}{3} \cdot- \sqrt{48}+\color{blue}{ \sqrt{48}} \cdot3+\color{blue}{ \sqrt{48}} \cdot- \sqrt{48} = \\ = 9- 12 \sqrt{3} + 12 \sqrt{3}-48 $$ |
| ③ | Simplify numerator and denominator |
| ④ | Multiply both numerator and denominator by -1. |
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