Do not solve the equation $\displaystyle x^2 - rx + s = 0$ for $s > 0, r > 2 \sqrt{s}$, but determine the number of real number of real solutions.
$
\begin{equation}
\begin{aligned}
x^2 - rx + s =& 0
&& \text{Given}
\\
\\
D =& b^2 - 4ac
&& \text{Discriminant Formula}
\\
\\
D =& (-r)^2 - 4(1)(s)
&& \text{Substitute the values}
\\
\\
D =& r - 4s
&& \text{Evaluate, suppose that } s =1
\\
\\
D =& 2 \sqrt{1} - 4(1)
&& \text{Substitute 1}
\\
\\
D <& 0
&& \text{Given the condition above, we can say that the function has no real solution since } D < 0
\end{aligned}
\end{equation}
$
Sunday, March 11, 2018
College Algebra, Chapter 1, 1.3, Section 1.3, Problem 72
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