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A portion of the quadratic formula proof is shown. Fill in the missing reason.

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

A. Multiply the fractions together on the right side of the equation
B. Subtract [tex]4 ac[/tex] on the right side of the equation
C. Add [tex]4 ac[/tex] to both sides of the equation



Answer :

GinnyAnswer
Certainly! Let's go through the proof step-by-step and fill in the missing reason.

Starting from:

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

Subtract [tex]\( c \)[/tex] from both sides:

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

Divide both sides by [tex]\( a \)[/tex]:

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

Complete the square on the left side and add [tex]\( \left(\frac{b}{2a}\right)^2 \)[/tex] to both sides:

4. [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)

Square [tex]\( \left(\frac{b}{2a}\right) \)[/tex] on the right side:

5. [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)

Find a common denominator on the right side of the equation:

6. [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)

Now, combine the fractions on the right side of the equation:

7. [tex]\( x^2 + \frac{b}{a}x + \left(\frac{b}{2a}\right)^2 = \frac{b^2 - 4ac}{4a^2} \)[/tex]

The missing reason is:

Combine the fractions together on the right side of the equation

So the completed proof is as follows:

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

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