Solve the equation by completing the square. $y ^ { 2 } + 5 y + 12 = 0$
A) $y = - \frac { 5 } { 2 } \pm$ $2 \sqrt { 3 }$
B) $y = \frac { 5 } { 2 } \pm \frac { \sqrt { 23 } i } { 2 }$
C) $y = - \frac { 5 } { 2 } \pm \frac { \sqrt { 23 } i } { 2 }$
D) $y = \frac { 5 } { 2 } \pm \frac { \sqrt { 38 } i } { 2 }$
E) $y = - \frac { 5 } { 2 } \pm \frac { \sqrt { 38 } i } { 2 }$

Question

Solve the equation by completing the square. $y ^ { 2 } + 5 y + 12 = 0$
A) $y = - \frac { 5 } { 2 } \pm$  $2 \sqrt { 3 }$
B) $y = \frac { 5 } { 2 } \pm \frac { \sqrt { 23 } i } { 2 }$
C) $y = - \frac { 5 } { 2 } \pm \frac { \sqrt { 23 } i } { 2 }$
D) $y = \frac { 5 } { 2 } \pm \frac { \sqrt { 38 } i } { 2 }$
E) $y = - \frac { 5 } { 2 } \pm \frac { \sqrt { 38 } i } { 2 }$

Quizplus · Accepted Answer

We start by grouping the first two terms and completing the square as shown below:
\begin{align*}
y^2 + 5y + 12 &=0\
\Rightarrow y^2 + 5y &= -12 \hspace{20pt} (	ext{subtracting 12 from both sides})\
\Rightarrow y^2 + 5y + \left(\frac{5}{2}ight)^2 &= -12 + \left(\frac{5}{2}ight)^2 \hspace{20pt} (	ext{adding and subtracting } \left(\frac{5}{2}ight)^2)\
\Rightarrow y^2 + 5y + \frac{25}{4} &= -12 + \frac{25}{4} \hspace{20pt} (	ext{simplifying})\
\Rightarrow \left(y+\frac{5}{2}ight)^2 &= \frac{1}{4} \
\Rightarrow \left(y+\frac{5}{2}ight)^2 &= \left(\frac{1}{2}ight)^2
\end{align*}
Taking the square root of both sides, we get:
\begin{align*}
y + \frac{5}{2} &= \pm \frac{1}{2} \
\Rightarrow y &= -\frac{5}{2} \pm \frac{1}{2} \
\Rightarrow y &= -\frac{5}{2} \pm \frac{\sqrt{1}}{2}i \
\Rightarrow y &= -\frac{5}{2} \pm \frac{\sqrt{1}\cdot \sqrt{23}}{2}i \
\Rightarrow y &= -\frac{5}{2} \pm \frac{\sqrt{23}}{2}i 
\end{align*}
Hence, the solution is $y = -\frac{5}{2} \pm \frac{\sqrt{23}}{2}i$, which corresponds to choice $\boxed{	extbf{(C)}}$.

Solve the Equation by Completing the Square y2+5y+12=0y ^ { 2 } + 5 y + 12 = 0y2+5y+12=0

Solve the Equation by Completing the Square $y ^ { 2 } + 5 y + 12 = 0$