Answer :
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]
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]
