Algebra - Quadratic Equations

Show Mobile Notice Show All Notes Hide All Notes

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 views in landscape mode. If your device is not in landscape mode many of the equations will run off the side of your device (should be able to scroll 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

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

\[{\left( {6t + 1} \right)^2} + 3 = 0\]

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 squared term on one side of the equation by itself with a coefficient of one. For this problem that gives,

\[{\left( {6t + 1} \right)^2} = - 3\] Show Step 2

Using the Square Root Property gives,

\[6t + 1 = \pm \sqrt { - 3} = \pm \sqrt 3 \,i\]

To finish this off all we need to do then is solve for \(t\) by subtracting 1 from both sides and then dividing by the 6. This gives,

\[\begin{align*}6t & = - 1 \pm \sqrt 3 \,i\\ t & = \frac{{ - 1 \pm \sqrt 3 \,i}}{6} = - \frac{1}{6} \pm \frac{{\sqrt 3 }}{6}i\end{align*}\]

Note that we did a little rewrite after dividing by the 6 to put the answer in a more standard form for complex numbers.

We then have the following two solutions : \(\require{bbox} \bbox[2pt,border:1px solid black]{{t = - \frac{1}{6} - \frac{{\sqrt 3 }}{6}i\,\,\,{\mbox{and }}t = - \frac{1}{6} + \frac{{\sqrt 3 }}{6}i}}\).