Question
Answer
$$\dfrac{\sqrt{13}}{\sqrt{13}+\sqrt{2}}=\dfrac{13-\sqrt{26}}{11}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{\sqrt{13}}{\sqrt{13}+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{\sqrt{13}}{\sqrt{13}+\sqrt{2}}\frac{\sqrt{13}-\sqrt{2}}{\sqrt{13}-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{13-\sqrt{26}}{13-\sqrt{26}+\sqrt{26}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{13-\sqrt{26}}{11}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{13}- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ \sqrt{13} } \cdot \left( \sqrt{13}- \sqrt{2}\right) = \color{blue}{ \sqrt{13}} \cdot \sqrt{13}+\color{blue}{ \sqrt{13}} \cdot- \sqrt{2} = \\ = 13- \sqrt{26} $$ Simplify denominator. $$ \color{blue}{ \left( \sqrt{13} + \sqrt{2}\right) } \cdot \left( \sqrt{13}- \sqrt{2}\right) = \color{blue}{ \sqrt{13}} \cdot \sqrt{13}+\color{blue}{ \sqrt{13}} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot \sqrt{13}+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 13- \sqrt{26} + \sqrt{26}-2 $$ |
| ③ | Simplify numerator and denominator |
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