Algebra - Quadratic Equations

Show General Notice Show Mobile Notice Show All Notes Hide All Notes

General Notice

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

Mobile Notice

You appear to be on a device with a "narrow" screen width (i.e. you are probably on a mobile phone). Due to the nature of the mathematics on this site it is best viewed in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (you should be able to scroll/swipe to see them) and some of the menu items will be cut off due to the narrow screen width.

Section 2.5 : Quadratic Equations - Part I

Back to Problem List

14. Use the Square Root Property to solve the equation.

\[{x^2} + 15 = 0\]

Show All Steps Hide All Steps

Start Solution

There really isn’t too much to this problem. Just recall that we need to get the variable on one side of the equation by itself with a coefficient of one. For this problem that gives,

\[{x^2} = - 15\] Show Step 2

Now all we need to do is use the Square Root Property to get,

\[x = \pm \sqrt { - 15} = \pm \sqrt {15} \,i\]

So we have the following two solutions : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \sqrt {15} \,i\,\,\,{\mbox{and }}x = \sqrt {15} \,i}}\] .

Do not get excited about complex solutions. They will happen fairly regularly when solving quadratic equations so we need to be able to deal with them.