College Algebra, Chapter 1, 1.3, Section 1.3, Problem 78

Suppose that the equation $kx^2 + 36x + k = 0$, find the values of $k$ that ensure that the given equation has exactly one solution.


$
\begin{equation}
\begin{aligned}

kx^2 + 36x + k =& 0
&& \text{Given}
\\
\\
D =& b^2 - 4ac
&& \text{Discriminant Formula}
\\
\\
0 =& b^2 - 4ac
&& \text{$D = 0$, for exactly one solution}
\\
\\
0 =& (36)^2 - 4(k)(k)
&& \text{Substitute the values}
\\
\\
0 =& (36)^2 - 4k^2
&& \text{Solve for } k
\\
\\
4k^2 =& 36^2
&& \text{Divide by 4, them take the square root}
\\
\\
k =& \pm \sqrt{324}
&&
\\
\\
k =& 18 \text{ and } k = -18
&&



\end{aligned}
\end{equation}
$