Question

Let quadratic equation be \(ax^2 + bx + c = 0\) then the sum of roots will be :

• A. \(\frac{b}{a}\)
• B. \(-\frac{b}{a}\)
• C. \(\frac{a}{b}\)
• D. \(-\frac{a}{b}\)

Hint:

Vieta's formulas apply to polynomial equations of any degree:
For any polynomial, the sum of the roots is always equal to:
\[ -\frac{\text{Coefficient of second highest term}}{\text{Coefficient of highest term}} \]
For a quadratic \(ax^2 + bx + c = 0\), this is simply \(-\frac{b}{a}\).

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
A quadratic equation is a second-degree polynomial equation of the form \(ax^2 + bx + c = 0\), where \(a \ne 0\).
The relationships between the coefficients of a polynomial equation and its roots are defined by Vieta's formulas.

Step 2: Detailed Explanation:
Let \(\alpha\) and \(\beta\) be the two roots of the quadratic equation \(ax^2 + bx + c = 0\).
According to the quadratic formula, the roots can be written as:
\[ \alpha = \frac{-b + \sqrt{b^2 - 4ac}}{2a} \]
\[ \beta = \frac{-b - \sqrt{b^2 - 4ac}}{2a} \]
Now, let us calculate the sum of these two roots:
\[ \alpha + \beta = \frac{-b + \sqrt{b^2 - 4ac}}{2a} + \frac{-b - \sqrt{b^2 - 4ac}}{2a} \]
Since they share a common denominator of \(2a\), we can combine the numerators:
\[ \alpha + \beta = \frac{-b + \sqrt{b^2 - 4ac} - b - \sqrt{b^2 - 4ac}}{2a} \]
The square root terms cancel out:
\[ \alpha + \beta = \frac{-2b}{2a} = -\frac{b}{a} \]
Thus, the sum of the roots is equal to \(-\frac{b}{a}\).
Additionally, the product of the roots is given by \(\alpha \beta = \frac{c}{a}\).

Step 3: Final Answer:
The correct option is (B).