Answer :
To fill in the missing statement, let's address the reasoning and follow the steps outlined in the proof of the quadratic formula.
Given the equation at the step before the missing statement:
[tex]\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]
The next step is to isolate [tex]\( x \)[/tex] by subtracting [tex]\(\frac{b}{2a}\)[/tex] from both sides of the equation.
Therefore, the missing statement should be:
[tex]\[ x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]
The final filled-in table will look like this:
\begin{tabular}{|c|c|}
\hline Statements & Reasons \\
\hline[tex]$x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=-\frac{4 a c}{4 a^2}+\frac{b^2}{4 a^2}$[/tex] & Find a common denominator on the right side of the equation \\
\hline \begin{tabular}{l}
[tex]$x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}$[/tex]
\end{tabular} & Add the fractions together on the right side of the equation \\
\hline \begin{tabular}{l}
[tex]$\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}$[/tex]
\end{tabular} & Rewrite the perfect square trinomial on the left side of the equation as a binomial squared \\
\hline \begin{tabular}{l}
[tex]$x+\frac{b}{2 a}= \pm \sqrt{\frac{b^2-4 a c}{4 a^2}}$[/tex]
\end{tabular} & Take the square root of both sides of the equation \\
\hline \begin{tabular}{l}
[tex]$x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}$[/tex]
\end{tabular} & Simplify the right side of the equation \\
\hline \begin{tabular}{l}
[tex]$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$[/tex]
\end{tabular} & Subtract [tex]$\left(\frac{ b }{2 a }\right)$[/tex] from both sides of the equation \\
\hline
\end{tabular}
Given the equation at the step before the missing statement:
[tex]\[ x + \frac{b}{2a} = \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]
The next step is to isolate [tex]\( x \)[/tex] by subtracting [tex]\(\frac{b}{2a}\)[/tex] from both sides of the equation.
Therefore, the missing statement should be:
[tex]\[ x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a} \][/tex]
The final filled-in table will look like this:
\begin{tabular}{|c|c|}
\hline Statements & Reasons \\
\hline[tex]$x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=-\frac{4 a c}{4 a^2}+\frac{b^2}{4 a^2}$[/tex] & Find a common denominator on the right side of the equation \\
\hline \begin{tabular}{l}
[tex]$x^2+\frac{b}{a} x+\left(\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}$[/tex]
\end{tabular} & Add the fractions together on the right side of the equation \\
\hline \begin{tabular}{l}
[tex]$\left(x+\frac{b}{2 a}\right)^2=\frac{b^2-4 a c}{4 a^2}$[/tex]
\end{tabular} & Rewrite the perfect square trinomial on the left side of the equation as a binomial squared \\
\hline \begin{tabular}{l}
[tex]$x+\frac{b}{2 a}= \pm \sqrt{\frac{b^2-4 a c}{4 a^2}}$[/tex]
\end{tabular} & Take the square root of both sides of the equation \\
\hline \begin{tabular}{l}
[tex]$x+\frac{b}{2 a}= \pm \frac{\sqrt{b^2-4 a c}}{2 a}$[/tex]
\end{tabular} & Simplify the right side of the equation \\
\hline \begin{tabular}{l}
[tex]$x = -\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$[/tex]
\end{tabular} & Subtract [tex]$\left(\frac{ b }{2 a }\right)$[/tex] from both sides of the equation \\
\hline
\end{tabular}
