Algebra - Quadratic Equations

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Section 2.5 : Quadratic Equations - Part I

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

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

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