Question
Answer
$$(t-4)^2+(21\cdot\dfrac{t}{4}-21)^2-1=t^2-8t+16+(21\cdot\dfrac{t}{4}-21)^2-1$$
Explanation
| $$ \begin{aligned}(t-4)^2+(21 \cdot \frac{t}{4}-21)^2-1& \xlongequal{ \color{blue}{ \text{\normalsize{ \htmlClass{explanationCircle explanationCircle1}{\textcircled {1}} } }}}t^2-8t+16+(21 \cdot \frac{t}{4}-21)^2-1\end{aligned} $$ | |
| ① | Find $ \left(t-4\right)^2 $ using formula. $$ (A - B)^2 = \color{blue}{A^2} - 2 \cdot A \cdot B + \color{red}{B^2} $$where $ A = \color{blue}{ t } $ and $ B = \color{red}{ 4 }$. $$ \begin{aligned}\left(t-4\right)^2 = \color{blue}{t^2} -2 \cdot t \cdot 4 + \color{red}{4^2} = t^2-8t+16\end{aligned} $$ |
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