Understanding the Quadratic Equation Problem

The question asks us to find a quadratic equation whose roots are \(a\) and \(b\). We are given a starting quadratic equation, \(x^2 - ax - bx + ab - c = 0\), and told that its roots are \(\alpha\) and \(\beta\).

To solve this, we need to connect the roots of the first equation (\(\alpha\), \(\beta\)) with its coefficients (which involve \(a\), \(b\), and \(c\)). Then, we will use the properties of the roots \(a\) and \(b\) to form the new quadratic equation.

Simplifying the Given Quadratic Equation

The given equation is \(x^2 - ax - bx + ab - c = 0\). We can rearrange this equation to group terms involving \(x\):

\(x^2 - (a+b)x + (ab - c) = 0\)

This is now in the standard form of a quadratic equation, \(Ax^2 + Bx + C = 0\), where \(A=1\), \(B=-(a+b)\), and \(C=ab-c\).

Applying Vieta's Formulas to the Given Equation

Vieta's formulas relate the roots of a polynomial to its coefficients. For a quadratic equation \(Ax^2 + Bx + C = 0\) with roots \(\alpha\) and \(\beta\), the formulas are:

• Sum of roots: \(\alpha + \beta = -\frac{B}{A}\)
• Product of roots: \(\alpha \beta = \frac{C}{A}\)

Using these formulas for our given equation \(x^2 - (a+b)x + (ab - c) = 0\) with roots \(\alpha\) and \(\beta\):

• Sum of roots: \(\alpha + \beta = -\frac{-(a+b)}{1} = a+b\)
• Product of roots: \(\alpha \beta = \frac{ab-c}{1} = ab-c\)

So, we have the following relationships:

• \(a+b = \alpha + \beta\)
• \(ab - c = \alpha \beta\)

From the second relationship, we can express \(ab\) in terms of \(\alpha\), \(\beta\), and \(c\):

• \(ab = \alpha \beta + c\)

Forming the New Quadratic Equation with Roots a and b

We want to find the quadratic equation whose roots are \(a\) and \(b\). A quadratic equation with roots \(r_1\) and \(r_2\) can be written as:

\(x^2 - (r_1+r_2)x + r_1r_2 = 0\)

In our case, the roots are \(a\) and \(b\). So, the sum of the roots for the new equation is \(a+b\), and the product of the roots is \(ab\).

Expressing Sum and Product in terms of α, β, and c

From our earlier findings using Vieta's formulas on the given equation, we know:

• Sum of the new roots: \(a+b = \alpha + \beta\)
• Product of the new roots: \(ab = \alpha \beta + c\)

Constructing the Desired Quadratic Equation

Now, substitute the sum and product of the new roots into the general form \(x^2 - (r_1+r_2)x + r_1r_2 = 0\):

\(x^2 - (a+b)x + ab = 0\)

Substitute \(a+b = \alpha + \beta\) and \(ab = \alpha \beta + c\):

\(x^2 - (\alpha + \beta)x + (\alpha \beta + c) = 0\)

Expanding the term with the sum of roots:

\(x^2 - \alpha x - \beta x + \alpha \beta + c = 0\)

This is the required quadratic equation whose roots are \(a\) and \(b\).

Comparing with Options

Let's compare our derived equation \(x^2 - \alpha x - \beta x + \alpha \beta + c = 0\) with the given options:

• Option 1: \(x^2 - \alpha x - \beta x + \alpha \beta + c = 0\)
• Option 2: \(x^2 - \alpha x - \beta x + \alpha \beta - c = 0\)
• Option 3: \(x^2 + \alpha x + \beta x + \alpha \beta + c = 0\)
• Option 4: \(x^2 + \alpha x + \beta x + \alpha \beta - c = 0\)

Our derived equation exactly matches Option 1.

Step-by-Step Derivation Summary

1. Identify the given quadratic equation: \(x^2 - ax - bx + ab - c = 0\).
2. Simplify the given equation into standard form: \(x^2 - (a+b)x + (ab - c) = 0\).
3. Apply Vieta's formulas to the given equation with roots \(\alpha\) and \(\beta\):
   • \(\alpha + \beta = a+b\)
   • \(\alpha \beta = ab - c\)
4. Rearrange the second Vieta's formula to find \(ab\): \(ab = \alpha \beta + c\).
5. Identify the roots of the desired equation: \(a\) and \(b\).
6. Determine the sum and product of the desired roots:
   • Sum: \(a+b\)
   • Product: \(ab\)
7. Substitute the expressions from steps 3 and 4 into the sum and product:
   • Sum: \(\alpha + \beta\)
   • Product: \(\alpha \beta + c\)
8. Form the quadratic equation using the sum and product of roots: \(x^2 - (\text{Sum})x + (\text{Product}) = 0\).
9. Substitute the expressions for sum and product: \(x^2 - (\alpha + \beta)x + (\alpha \beta + c) = 0\).
10. Expand the equation: \(x^2 - \alpha x - \beta x + \alpha \beta + c = 0\).
11. Compare with the options to find the matching equation.

Revision Table: Key Concepts

Additional Information: Generalizing Vieta's Formulas

Vieta's formulas can be generalized to polynomials of any degree. For a polynomial equation \(a_n x^n + a_{n-1} x^{n-1} + \dots + a_1 x + a_0 = 0\) with roots \(r_1, r_2, \dots, r_n\), the formulas relate the coefficients to the elementary symmetric polynomials of the roots.

• Sum of roots: \(r_1 + r_2 + \dots + r_n = -\frac{a_{n-1}}{a_n}\)
• Sum of products of roots taken two at a time: \(\sum_{i<j} r_i r_j = \frac{a_{n-2}}{a_n}\)
• ...
• Product of roots: \(r_1 r_2 \dots r_n = (-1)^n \frac{a_0}{a_n}\)

In the context of quadratic equations (\(n=2\), \(a_2 x^2 + a_1 x + a_0 = 0\)), this simplifies to:

• Sum of roots (\(r_1+r_2\)): \(-\frac{a_1}{a_2}\)
• Product of roots (\(r_1 r_2\)): \(\frac{a_0}{a_2}\)

These are the formulas we used in this problem.