Monday, January 29, 2018

College Algebra, Chapter 1, 1.6, Section 1.6, Problem 90

The equation $\displaystyle T = \frac{600,000}{x^2 + 300}$ represents the temperature in $^\circ C$ at a distance $x$ refers from the center of the fire in the vicinity of a bonfire. At what range of distance from the fire's center was the temperature less than $50^\circ C$

$
\begin{equation}
\begin{aligned}
\frac{600,000}{x^2 + 300} &< 500 && \text{model}\\
\\
\frac{600,000}{x^2 + 300} -500 &< 0 && \text{Subtract 500}\\
\\
\frac{600,000 - 500 x^2 - 150,000}{x^2 + 300} & < 0 && \text{Common Denominators}\\
\\
\frac{-500x^2 + 450,000}{x^2 + 300} & < 0 && \text{Simplify the numerator}\\
\\
\frac{-500(x^2 - 900)}{x^2 + 300} &< 0 && \text{Factor out -500}\\
\\
\frac{x^2 - 900}{x^2 + 300} & > 0 && \text{Divide both sides by -500}\\
\\
\frac{(x+30)(x-30)}{x^2 + 300} & > 0 && \text{Difference of squares}
\end{aligned}
\end{equation}
$


The real solution will be $x - 30 > 0$. It shows that if the distance of the man from the fire is greater than 30m, the temperature will be less than $500^\circ C$

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