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      <text>1.2.5 (Completing the square - general quadratic)</text>
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      <text><![CDATA[<h1>Solving a general quadratic</h1>

<p>It is possible to use the method of completing the square to solve a general quadratic \(y = ax^2 + bx + c\). So far, we have encountered examples where the leading coefficient of the quadratic is \(1\). However, it is not much more difficult to deal with a quadratic where \(a \not=1\). </p>
<br>
<h3>Example</h3>

<p>Solve the quadratic \(2x^2 + 12x + 4=0\) by completing the square.</p>
<br>
<h3>Solution</h3>

<p>We can immediately deal with the leading coefficient by dividing all the terms through by \(2\) to obtain
\[
x^2 + 6x + 2 = 0.
\]
Now we have a similar problem to those we have seen previously. We complete the square by adding and subtracting \(\frac{b}{2}^2 = 3^2 = 9\), (or equivalently adding \(\frac{b}{2}^2\) to both the LHS and RHS).
\[
\begin{eqnarray*}
x^2 + 6x + 2 + 9 - 9 &amp;= x^2 + 6x + 9 + 2 - 9 
\\
&amp;= (x+3)^2 -7 
\\ 
&amp;= 0.
\end{eqnarray*}
\]
At this point we can add \(7\) to both sides, then proceed by taking square roots.
\[
(x+3)^2 = 7 \implies x+ 3 = \pm \sqrt{7} \implies x = -3 \pm \sqrt{7}.
\]
Therefore the solutions to the quadratic are \(x = -3 + \sqrt{7}\) and \(x = -3 - \sqrt{7}.\)
</p>]]></text>
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