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### Section 2.6 : Quadratic Equations - Part II

8. Solve the following quadratic equation by completing the square.

$2{x^2} + 5x + 3 = 0$

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First, let’s get the equation put into the form where all the variables are on one side, with a coefficient of one on the $${x^2}$$, and the number is on the other side. Doing this gives,

\begin{align*}{x^2} + \frac{5}{2}x + \frac{3}{2} & = 0\\ {x^2} + \frac{5}{2}x & = - \frac{3}{2}\end{align*} Show Step 2

We 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,

${\left( {\frac{{\frac{5}{2}}}{2}} \right)^2} = {\left( {\frac{5}{4}} \right)^2} = \frac{{25}}{{16}}$ Show Step 3

At 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*}{x^2} + \frac{5}{2}x \,{\color{Red} + \frac{{25}}{{16}}} & = - \frac{3}{2} \,{\color{Red} + \frac{{25}}{{16}}}\\ {\left( {x + \frac{5}{4}} \right)^2} & = \frac{1}{{16}}\end{align*} Show Step 4

Now 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*}x + \frac{5}{4} & = \pm \sqrt {\frac{1}{{16}}} = \pm \frac{1}{4}\\ x & = - \frac{5}{4} \pm \frac{1}{4}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = - \frac{5}{4} - \frac{1}{4} = - \frac{3}{2},\,\,\,\,\,\,\,\,x = - \frac{5}{4} + \frac{1}{4} = - 1\end{align*}

The two solutions to this equation are then : $\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{3}{2}\,\,{\mbox{and }}x = - 1}}$ .