Question
Answer
$$\dfrac{-8}{2+\sqrt{2}}=\dfrac{-16+8\sqrt{2}}{2}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}\frac{-8}{2+\sqrt{2}}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}\frac{-8}{2+\sqrt{2}}\frac{2-\sqrt{2}}{2-\sqrt{2}} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}\frac{-16+8\sqrt{2}}{4-2\sqrt{2}+2\sqrt{2}-2} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}\frac{-16+8\sqrt{2}}{2}\end{aligned} $$ | |
| ① | Multiply the numerator and denominator by the conjugate of the denominator . $$\color{blue}{ 2- \sqrt{2}} $$. |
| ② | Multiply in a numerator. $$ \color{blue}{ -8 } \cdot \left( 2- \sqrt{2}\right) = \color{blue}{-8} \cdot2\color{blue}{-8} \cdot- \sqrt{2} = \\ = -16 + 8 \sqrt{2} $$ Simplify denominator. $$ \color{blue}{ \left( 2 + \sqrt{2}\right) } \cdot \left( 2- \sqrt{2}\right) = \color{blue}{2} \cdot2+\color{blue}{2} \cdot- \sqrt{2}+\color{blue}{ \sqrt{2}} \cdot2+\color{blue}{ \sqrt{2}} \cdot- \sqrt{2} = \\ = 4- 2 \sqrt{2} + 2 \sqrt{2}-2 $$ |
| ③ | Simplify numerator and denominator |
This page was created using
Rationalize Denominator Calculator