To understand why [tex]\(\frac{b^2}{4a^2}\)[/tex] is not added but rather subtracted from the left side of the equation, we need to carefully look at the process of completing the square in the quadratic equation:
1. Start with the quadratic equation in standard form:
[tex]\[
ax^2 + bx + c = 0
\][/tex]
2. Move the constant term [tex]\(c\)[/tex] to the left side:
[tex]\[
-c = -ax^2 - bx
\][/tex]
3. Factor out the coefficient [tex]\(a\)[/tex] from the terms on the right:
[tex]\[
-c = -a \left( x^2 + \frac{b}{a}x \right)
\][/tex]
4. To complete the square, find the term to add inside the parentheses to make it a perfect square trinomial. This term is [tex]\(\left( \frac{b}{2a} \right)^2\)[/tex]:
[tex]\[
\left( \frac{b}{2a} \right)^2 = \frac{b^2}{4a^2}
\][/tex]
5. Add this term inside the parentheses and balance it by subtracting it on the left side. By adding this term inside the parentheses, you modify the equation to:
[tex]\[
-c - a \left( \frac{b^2}{4a^2} \right) = -a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right)
\][/tex]
Since you added [tex]\(\frac{b^2}{4a^2}\)[/tex] inside the parentheses on the right (multiplied by [tex]\(-a\)[/tex]), you need to subtract the equivalent term outside:
[tex]\[
-c + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} = -a \left( x^2 + \frac{b}{a}x + \frac{b^2}{4a^2} \right)
\][/tex]
6. Simplify the expressions:
[tex]\[
-c + \frac{b^2}{4a^2} - a \left( x^2 + \frac{b}{a}x - \frac{b^2}{4a^2} \right)
\][/tex]
In step 5, the term [tex]\(\frac{b^2}{4a^2}\)[/tex] is added inside the parentheses on the right side of the equation. To maintain the balance of the equation, you must subtract the same term from the left side of the equation. This ensures that both sides of the equation remain equal after completing the square:
[tex]\[
\boxed{\text{The term } \frac{b^2}{4 a^2} \text{ is added to the right side of the equation, so it needs to be subtracted from the left side of the equation to balance the sides of the equation.}}
\][/tex]
This explanation clarifies why [tex]\(\frac{b^2}{4a^2}\)[/tex] is not added but instead is subtracted from the left side in the last step; thereby maintaining the equality of both sides of the equation.
