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The first few steps in deriving the quadratic formula are shown.




$-c=a x^2+b x$		Use the subtraction property of equality.

$-c=a\left(x^2+\frac{b}{a} x\right)$		Factor out $a$.

$\left(\frac{b}{2 a}\right)^2=\frac{b^2}{4 a^2}$		Find half of the $b$ value and square it to determine the constant of the perfect square trinomial.

$-c+\frac{b^2}{4 a}=a\left(x^2+\frac{b}{a} x+\frac{b^2}{4 a^2}\right)$






Which best explains why $\frac{b^2}{4 a^2}$ is not added to the left side of the equation in the last step shown in the table?


AThe term $b^2$ is added to the right side of the equation, so it needs to be subtracted from the left side of the equation to balance the sides of the equation.
BThe distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation.
CThe term $\frac{b^2}{4 a^2}$ needs to be converted so it has a common denominator before adding it to the left side of the equation to balance the equation.
DThe square root of the term needs to be found before adding it to the left side of the equation.

Verified Answer :

  • We start with the equation \(ax^2 + bx = -c\) and factor out \(a\).
  • To complete the square, we add \(\frac{b^2}{4a^2}\) inside the parentheses.
  • Due to the distributive property, the actual value added to the right side is \(a \cdot \frac{b^2}{4a^2} = \frac{b^2}{4a}\).
  • To keep the equation balanced, \(\frac{b^2}{4a}\) must be added to the left side. The correct explanation is \(\boxed{\text{The distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation.}}\)

Explanation

  1. Initial Steps of the Derivation

To derive the quadratic formula starting from the standard form \(ax^2 + bx + c = 0\), we first isolate the quadratic and linear terms by subtracting \(c\) from both sides:
\[ax^2 + bx = -c.\]
Next, we factor out the coefficient \(a\) from the quadratic and linear terms on the left side:
\[a\left(x^2 + \frac{b}{a}x\right) = -c.\]

  1. Applying the Distributive Property

To complete the square within the parentheses, we take half of the coefficient of \(x\), which is \(\frac{b}{a}\), divide by 2 to get \(\frac{b}{2a}\), and then square this result to obtain \(\frac{b^2}{4a^2}\). We then add this value inside the parentheses:
\[a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right).\]
However, adding this term inside the parentheses effectively changes the value of the right side. According to the distributive property, multiplying this new term by the factor \(a\) outside the parentheses gives:
\[a \times \frac{b^2}{4a^2} = \frac{b^2}{4a}.\]
This means the expression on the right side has increased by exactly \(\frac{b^2}{4a}\).

  1. Balancing the Equation

To maintain the equality of the equation, any value added to one side must also be added to the other. Since adding \(\frac{b^2}{4a^2}\) inside the parentheses on the right side results in an effective increase of \(\frac{b^2}{4a}\), we must add \(\frac{b^2}{4a}\) to the left side. This leads to the equation:
\[-c + \frac{b^2}{4a} = a\left(x^2 + \frac{b}{a}x + \frac{b^2}{4a^2}\right).\]
Therefore, the reason \(\frac{b^2}{4a^2}\) is not added directly to the left side is that the distributive property must be applied to account for the factor \(a\) outside the parentheses.

  1. Conclusion and Final Answer Selection

Comparing our analysis with the provided options, the statement that best explains why \(\frac{b^2}{4a^2}\) is not added to the left side is:
"The distributive property needs to be applied to determine the value to add to the left side of the equation to balance the sides of the equation."
This aligns with our finding that the effective increase on the right side is \(\frac{b^2}{4a}\) due to the distributive property, necessitating that same value be added to the left side. The final answer is the second option.

Examples

Understanding how to maintain equality in an equation when working with grouped terms is a fundamental skill in algebra. For example, when balancing a chemical equation or calculating compound interest, you often manipulate expressions within brackets. Applying the distributive property correctly ensures that your mathematical models accurately represent the physical or financial processes you are studying. This step-by-step approach prevents errors and ensures the integrity of your calculations.

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