I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 2.6 : Quadratic Equations - Part II
5. Solve the following quadratic equation by completing the square.
\[{v^2} + 8v - 9 = 0\]Show All Steps Hide All Steps
Start SolutionFirst, let’s get the equation put into the form where all the variables are on one side and the number is on the other side. Doing this gives,
\[{v^2} + 8v = 9\] Show Step 2We can now complete the square on the expression on the left side of the equation.
The number that we’ll need to do this is,
\[{\left( {\frac{8}{2}} \right)^2} = {\left( 4 \right)^2} = 16\] Show Step 3At this point we need to recall that we have an equation here and what we do to one side of the equation we also need to do the other. In other words, don’t forget to add the number from the previous step to both sides of the equation from Step 1.
\[\require{color}\begin{align*}{v^2} + 8v \,{\color{Red}+ 16} & = 9 \,{\color{Red}+ 16}\\ {\left( {v + 4} \right)^2} & = 25\end{align*}\] Show Step 4Now all we need to do to finish solving the equation is to use the Square Root Property on the equation from the previous step. Doing this gives,
\[\begin{align*}v + 4 & = \pm \sqrt {25} = \pm 5\\ v & = - 4 \pm 5\hspace{0.25in} \Rightarrow \hspace{0.25in}v = - 4 - 5 = - 9,\,\,\,\,\,v = - 4 + 5 = 1\end{align*}\]The two solutions to this equation are then : \[\require{bbox} \bbox[2pt,border:1px solid black]{{v = - 9\,\,{\mbox{and }}v = 1}}\] .