Question
Answer
$$(9+i)sqrt\dfrac{-65-137i}{82}=\dfrac{-137i^2qrst-1298iqrst-585qrst}{82}$$
Explanation
Tap the blue circles to see an explanation.
| $$ \begin{aligned}(9+i)sqrt\frac{-65-137i}{82}& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}(9s+is)qrt\frac{-65-137i}{82} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}(9qs+iqs)rt\frac{-65-137i}{82} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle3}{\textcircled {3}} } }}}(9qrs+iqrs)t\frac{-65-137i}{82} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle4}{\textcircled {4}} } }}}(9qrst+iqrst)\frac{-65-137i}{82} \xlongequal{ } \\[1 em] & \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle5}{\textcircled {5}} } }}}\frac{-137i^2qrst-1298iqrst-585qrst}{82}\end{aligned} $$ | |
| ① | $$ \left( \color{blue}{9+i}\right) \cdot s = 9s+is $$ |
| ② | $$ \left( \color{blue}{9s+is}\right) \cdot q = 9qs+iqs $$ |
| ③ | $$ \left( \color{blue}{9qs+iqs}\right) \cdot r = 9qrs+iqrs $$ |
| ④ | $$ \left( \color{blue}{9qrs+iqrs}\right) \cdot t = 9qrst+iqrst $$ |
| ⑤ | Multiply $9qrst+iqrst$ by $ \dfrac{-65-137i}{82} $ to get $ \dfrac{-137i^2qrst-1298iqrst-585qrst}{82} $. Step 1: Write $ 9qrst+iqrst $ as a fraction by putting $ \color{red}{1} $ in the denominator. Step 2: Multiply numerators and denominators. Step 3: Simplify numerator and denominator. $$ \begin{aligned} 9qrst+iqrst \cdot \frac{-65-137i}{82} & \xlongequal{\text{Step 1}} \frac{9qrst+iqrst}{\color{red}{1}} \cdot \frac{-65-137i}{82} = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \left( 9qrst+iqrst \right) \cdot \left( -65-137i \right) }{ 1 \cdot 82 } \xlongequal{\text{Step 3}} \frac{ -585qrst-1233iqrst-65iqrst-137i^2qrst }{ 82 } = \\[1ex] &= \frac{-137i^2qrst-1298iqrst-585qrst}{82} \end{aligned} $$ |
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