Understanding the Quadratic Equation

The question provides a quadratic equation in the form \( (a + b) x^2 - (a + b + c) x + k = 0 \), where a, b, and c are real numbers. We are asked to find the roots of this equation under a specific condition.

Substituting k=c into the Equation

The condition given is that \( k = c \). Substituting this into the original quadratic equation, we get:

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

This is a quadratic equation of the form \( Ax^2 + Bx + C = 0 \), where:

• \( A = a + b \)
• \( B = -(a + b + c) \)
• \( C = c \)

We need to find the values of \( x \) that satisfy this equation, which are called the roots.

Finding the Roots of the Equation

There are several ways to find the roots of a quadratic equation. We can use the quadratic formula, factor the equation, or if we suspect a simple root, we can test it directly. Let's examine the form of the equation: \( (a + b) x^2 - (a + b + c) x + c = 0 \). Notice the coefficients.

Testing a Potential Root (x=1)

Sometimes, simple integer roots like 1 or -1 are present. Let's test if \( x = 1 \) is a root of the equation. Substitute \( x = 1 \) into the equation:

\( (a + b) (1)^2 - (a + b + c) (1) + c \)

\( = (a + b) - (a + b + c) + c \)

\( = a + b - a - b - c + c \)

\( = (a - a) + (b - b) + (-c + c) \)

\( = 0 + 0 + 0 = 0 \)

Since substituting \( x = 1 \) results in 0, \( x = 1 \) is indeed one of the roots of the equation \( (a + b) x^2 - (a + b + c) x + c = 0 \).

Using Root Properties (Vieta's Formulas)

For a quadratic equation \( Ax^2 + Bx + C = 0 \) with roots \( x_1 \) and \( x_2 \), Vieta's formulas state that:

• Sum of roots: \( x_1 + x_2 = -\frac{B}{A} \)
• Product of roots: \( x_1 x_2 = \frac{C}{A} \)

In our equation, \( (a + b) x^2 - (a + b + c) x + c = 0 \), we have \( A = a+b \), \( B = -(a+b+c) \), and \( C = c \). We know one root is \( x_1 = 1 \). Let the other root be \( x_2 \).

Using the product of roots formula:

\( x_1 x_2 = \frac{C}{A} \)

\( 1 \cdot x_2 = \frac{c}{a+b} \)

\( x_2 = \frac{c}{a+b} \)

Alternatively, using the sum of roots formula:

\( x_1 + x_2 = -\frac{B}{A} \)

\( 1 + x_2 = -\frac{-(a+b+c)}{a+b} \)

\( 1 + x_2 = \frac{a+b+c}{a+b} \)

\( x_2 = \frac{a+b+c}{a+b} - 1 \)

\( x_2 = \frac{a+b+c - (a+b)}{a+b} \)

\( x_2 = \frac{a+b+c - a - b}{a+b} \)

\( x_2 = \frac{c}{a+b} \)

Both methods confirm that the second root is \( \frac{c}{a+b} \).

The Roots of the Equation

Therefore, the roots of the quadratic equation \( (a + b) x^2 - (a + b + c) x + k = 0 \) when \( k = c \) are \( 1 \) and \( \frac{c}{a+b} \).

Revision Table: Quadratic Equations and Roots

Additional Information: Vieta's Formulas Explained

Vieta's formulas provide a relationship between the coefficients of a polynomial equation and the sums and products of its roots. For a quadratic equation \( Ax^2 + Bx + C = 0 \), the formulas are particularly simple and useful.

The sum of the roots \( x_1 + x_2 \) is equal to the negative of the coefficient of the \( x \) term (\( B \)) divided by the coefficient of the \( x^2 \) term (\( A \)). This property was used in the solution to verify the second root.

The product of the roots \( x_1 x_2 \) is equal to the constant term (\( C \)) divided by the coefficient of the \( x^2 \) term (\( A \)). This property was also used to directly calculate the second root knowing the first one.

These formulas are derived from the fact that if \( x_1 \) and \( x_2 \) are roots, the quadratic equation can be factored as \( A(x - x_1)(x - x_2) = 0 \). Expanding this gives \( A(x^2 - (x_1+x_2)x + x_1x_2) = 0 \), which is \( Ax^2 - A(x_1+x_2)x + A(x_1x_2) = 0 \). Comparing this to \( Ax^2 + Bx + C = 0 \), we see that \( B = -A(x_1+x_2) \) and \( C = A(x_1x_2) \), leading directly to Vieta's formulas.