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Question

Answer

$$4x(x+\dfrac{3}{2}-i^{1/2})(x+\dfrac{3}{2}-i^{1/2})=4x(\dfrac{2x+3}{2}-i^{1/2})(\dfrac{2x+3}{2}-i^{1/2})$$

Explanation

Tap the blue circles to see an explanation.

$$ \begin{aligned}4x(x+\frac{3}{2}-i^{1/2})(x+\frac{3}{2}-i^{1/2})& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} \htmlClass{explanationCircle explanationCircle2}{\textcircled {2}} } }}}4x(\frac{2x+3}{2}-i^{1/2})(\frac{2x+3}{2}-i^{1/2})\end{aligned} $$
①

Add $x$ and $ \dfrac{3}{2} $ to get $ \dfrac{ \color{purple}{ 2x+3 } }{ 2 }$.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} x+ \frac{3}{2} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} + \frac{3}{2} = \frac{ x \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} + \frac{ 3 }{ 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x } }{ 2 } + \frac{ \color{purple}{ 3 } }{ 2 }=\frac{ \color{purple}{ 2x+3 } }{ 2 } \end{aligned} $$
②

Add $x$ and $ \dfrac{3}{2} $ to get $ \dfrac{ \color{purple}{ 2x+3 } }{ 2 }$.

Step 1: Write $ x $ as a fraction by putting $ \color{red}{ 1 } $ in the denominator.

Step 2: To add raitonal expressions, both fractions must have the same denominator.
We can create a common denominator by multiplying the first fraction by $\color{blue}{ 2 }$.

$$ \begin{aligned} x+ \frac{3}{2} & \xlongequal{\text{Step 1}} \frac{x}{\color{red}{1}} + \frac{3}{2} = \frac{ x \cdot \color{blue}{ 2 }}{ 1 \cdot \color{blue}{ 2 }} + \frac{ 3 }{ 2 } = \\[1ex] & \xlongequal{\text{Step 2}} \frac{ \color{purple}{ 2x } }{ 2 } + \frac{ \color{purple}{ 3 } }{ 2 }=\frac{ \color{purple}{ 2x+3 } }{ 2 } \end{aligned} $$
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