Finding the Roots of a Quadratic Equation

The problem asks us to find the roots of the given quadratic equation:

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

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

• \(A = (a + b + c)\)
• \(B = -(2a + 2b)\)
• \(C = (a + b - c)\)

We are given that \(a\), \(b\), and \(c\) are real and distinct.

Checking for Simple Roots

Sometimes, quadratic equations with structured coefficients have simple roots like \(x = 1\) or \(x = -1\). Let's test if \(x = 1\) is a root by substituting \(x=1\) into the equation:

Left Hand Side (LHS) \( = (a + b + c) (1)^2 - (2a + 2b) (1) + (a + b - c)\)

LHS \( = (a + b + c) - (2a + 2b) + (a + b - c)\)

LHS \( = a + b + c - 2a - 2b + a + b - c\)

Now, let's group the terms with \(a\), \(b\), and \(c\):

LHS \( = (a - 2a + a) + (b - 2b + b) + (c - c)\)

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

LHS \( = 0\)

Since substituting \(x=1\) results in 0, \(x=1\) is indeed one of the roots of the quadratic equation.

Using the Properties of Roots

For a quadratic equation \(Ax^2 + Bx + C = 0\), if the roots are \(r_1\) and \(r_2\), we know that:

• Sum of roots: \(r_1 + r_2 = -\frac{B}{A}\)
• Product of roots: \(r_1 \cdot r_2 = \frac{C}{A}\)

We have found one root, let's say \(r_1 = 1\). We can use either the sum or product of roots property to find the other root, \(r_2\).

Method 1: Using the Sum of Roots

The sum of the roots is \(r_1 + r_2 = -\frac{B}{A}\).

Substitute the values of \(A\) and \(B\):

\(1 + r_2 = -\frac{-(2a + 2b)}{(a + b + c)}\)

\(1 + r_2 = \frac{2a + 2b}{a + b + c}\)

Now, solve for \(r_2\):

\(r_2 = \frac{2a + 2b}{a + b + c} - 1\)

To subtract 1, we can write 1 as \(\frac{a+b+c}{a+b+c}\):

\(r_2 = \frac{2a + 2b}{a + b + c} - \frac{a + b + c}{a + b + c}\)

\(r_2 = \frac{(2a + 2b) - (a + b + c)}{a + b + c}\)

\(r_2 = \frac{2a + 2b - a - b - c}{a + b + c}\)

\(r_2 = \frac{(2a - a) + (2b - b) - c}{a + b + c}\)

\(r_2 = \frac{a + b - c}{a + b + c}\)

So, the other root is \(\frac{a + b - c}{a + b + c}\).

Method 2: Using the Product of Roots

The product of the roots is \(r_1 \cdot r_2 = \frac{C}{A}\).

Substitute the values of \(A\) and \(C\):

\(1 \cdot r_2 = \frac{(a + b - c)}{(a + b + c)}\)

\(r_2 = \frac{a + b - c}{a + b + c}\)

Both methods give the same result for the second root.

Therefore, the roots of the equation are \(1\) and \(\frac{a + b - c}{a + b + c}\).

Summary of Roots

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

Revision Table: Key Concepts for Quadratic Equations

Additional Information: Solving Quadratic Equations

Understanding how to find the roots of quadratic equations is a fundamental skill in algebra. There are several methods to solve a quadratic equation \(Ax^2 + Bx + C = 0\):

1. Factoring: If the quadratic expression \(Ax^2 + Bx + C\) can be factored into the product of two linear factors, say \((px + q)(rx + s)\), then the roots are found by setting each factor to zero: \(px + q = 0\) and \(rx + s = 0\). This gives \(x = -q/p\) and \(x = -s/r\).
2. Using the Quadratic Formula: The quadratic formula provides a direct way to find the roots for any quadratic equation, regardless of whether it can be easily factored. The formula is \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\). The term \(B^2 - 4AC\) is called the discriminant, which tells us about the nature of the roots (real, distinct, equal, complex).
3. Completing the Square: This method involves manipulating the equation algebraically to form a perfect square trinomial on one side, which then allows you to take the square root of both sides to solve for \(x\).
4. Testing Simple Roots: For equations with integer coefficients or coefficients with a clear pattern, testing simple values like 0, 1, -1, 2, -2 can sometimes quickly reveal a root. If \(x=k\) is a root, then \((x-k)\) is a factor of the quadratic expression.

In this specific problem, testing \(x=1\) was a quick way to find one root because the sum of the coefficients \(A + B + C\) turned out to be zero. For any polynomial \(P(x)\), if the sum of its coefficients is zero, then \(x=1\) is a root of \(P(x)=0\).