Fill in the missing reason in the proof of the quadratic formula.

\begin{tabular}{|l|l|}
\hline \multicolumn{1}{|c|}{Statements} & \multicolumn{1}{|c|}{Reasons} \\
\hline
[tex]$a x^2 + b x + c = 0$[/tex] & Given \\
\hline
[tex]$a x^2 + b x = -c$[/tex] & Subtract [tex]$c$[/tex] from both sides of the equation \\
\hline
[tex]$x^2 + \frac{b}{a} x = -\frac{c}{a}$[/tex] & Divide both sides of the equation by [tex]$a$[/tex] \\
\hline
[tex]$x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2$[/tex] & Complete the square and add [tex]$\left(\frac{b}{2a}\right)^2$[/tex] to both sides \\
\hline
[tex]$x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2}$[/tex] & Square [tex]$\left(\frac{b}{2a}\right)$[/tex] on the right side of the equation \\
\hline
[tex]$x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2}$[/tex] & Find a common denominator on the right side of the equation \\
\hline
[tex]$x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$[/tex] & Add the fractions together on the right side of the equation \\
\hline
[tex]$\left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2}$[/tex] & Factor the left side of the equation \\
\hline
\end{tabular}



Answer :

Certainly! Let's walk through the process step-by-step in detail.

We start with the general quadratic equation:

1. Statement: [tex]\( ax^2 + bx + c = 0 \)[/tex]
Reason: Given

2. Statement: [tex]\( ax^2 + bx = -c \)[/tex]
Reason: Subtract [tex]\( c \)[/tex] from both sides of the equation

3. Statement: [tex]\( x^2 + \frac{b}{a}x = -\frac{c}{a} \)[/tex]
Reason: Divide both sides of the equation by [tex]\( a \)[/tex]

4. Statement: [tex]\( x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \left(\frac{b}{2a}\right)^2 \)[/tex]
Reason: Complete the square by adding [tex]\(\left(\frac{b}{2a}\right)^2\)[/tex] to both sides

5. Statement: [tex]\( x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = -\frac{c}{a} + \frac{b^2}{4a^2} \)[/tex]
Reason: Square [tex]\(\left(\frac{b}{2a}\right)^2\)[/tex] on the right side of the equation

6. Statement: [tex]\( x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = -\frac{4ac}{4a^2} + \frac{b^2}{4a^2} \)[/tex]
Reason: Find a common denominator on the right side of the equation

7. Statement: [tex]\( x^2 + \frac{b}{a} x + \left(\frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)[/tex]
Reason: Add the fractions together on the right side of the equation

8. Statement: [tex]\( \left(x + \frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)[/tex]
Reason: Rewrite the left side as a square of the binomial

By following these steps, we transform the quadratic equation into a form that allows us to recognize the left side as a perfect square trinomial, making it easier to further solve the equation using the quadratic formula.

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