Question
Answer
$$\dfrac{2\sqrt{15}+2\sqrt{6}}{\sqrt{6}\cdot0-\sqrt{24}}=-\dfrac{\sqrt{10}+2}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{2\sqrt{15}+2\sqrt{6}}{\sqrt{6}\cdot0-\sqrt{24}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{2\sqrt{15}+2\sqrt{6}}{-\sqrt{24}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{2\sqrt{15}+2\sqrt{6}}{-\sqrt{24}}\frac{\sqrt{24}}{\sqrt{24}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{12\sqrt{10}+24}{-24} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}\frac{\sqrt{10}+2}{-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}-\frac{\sqrt{10}+2}{2}\end{aligned} $$ | |
| ① | Simplify numerator and denominator |
| ② | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ \sqrt{24}} $$. |
| ③ | Multiply in a numerator. $$ \color{blue}{ \left( 2 \sqrt{15} + 2 \sqrt{6}\right) } \cdot \sqrt{24} = \color{blue}{ 2 \sqrt{15}} \cdot \sqrt{24}+\color{blue}{ 2 \sqrt{6}} \cdot \sqrt{24} = \\ = 12 \sqrt{10} + 24 $$ Simplify denominator. $$ \color{blue}{ - \sqrt{24} } \cdot \sqrt{24} = -24 $$ |
| ④ | Divide both numerator and denominator by 12. |
| ⑤ | Place a negative sign in front of a fraction. |
This page was created using
Rationalize Denominator Calculator