Understanding the Quadratic Equation

The given quadratic equation is:

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

where \(a\), \(b\), and \(c\) are real and distinct numbers.

Let the coefficients of the quadratic equation be \(A\), \(B\), and \(C\), so the equation is in the form \(Ax^2 + Bx + C = 0\). Here:

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

For this to be a quadratic equation, the coefficient of \(x^2\) must be non-zero, i.e., \(A = a + b + c \neq 0\).

Finding the Roots of the Equation

Let's try to find the roots. A common technique is to test simple values for \(x\). Let's see what happens if we substitute \(x=1\) into the equation:

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

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

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

Group the terms with \(a\), \(b\), and \(c\):

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

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

Since substituting \(x=1\) makes the equation true, \(x=1\) is always one of the roots of this quadratic equation, provided \(a+b+c \neq 0\).

Let the roots be \(x_1\) and \(x_2\). We know \(x_1 = 1\). For a quadratic equation \(Ax^2 + Bx + C = 0\), the product of the roots is given by \(x_1 x_2 = \frac{C}{A}\).

Using this formula:

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

So, the other root is \(x_2 = \frac{a + b - c}{a + b + c}\).

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

Analyzing Statement 1: Roots when a, b, c are Positive

Statement 1 says: One of the roots of the equation is always less than 1 if \(a\), \(b\), and \(c\) are all positive.

We know the roots are \(x_1=1\) and \(x_2 = \frac{a+b-c}{a+b+c}\). Since \(x_1=1\) is not less than 1, this statement implies that the other root, \(x_2\), must be less than 1 when \(a, b, c > 0\).

Let's assume \(a > 0\), \(b > 0\), and \(c > 0\). We need to check if \(x_2 < 1\):

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

Since \(a, b, c\) are all positive, their sum \(a+b+c\) is also positive. We can multiply both sides of the inequality by \((a+b+c)\) without changing the direction of the inequality sign:

\(a+b-c < 1 \cdot (a+b+c)\)

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

Subtract \((a+b)\) from both sides:

\(-c < c\)

Add \(c\) to both sides:

\(0 < 2c\)

Divide by 2:

\(0 < c\)

This result \(0 < c\) is true under the condition that \(c\) is positive. Since Statement 1 assumes \(a, b, c\) are all positive, \(c > 0\) is true.

Therefore, if \(a, b, c\) are all positive, the other root \(x_2 = \frac{a+b-c}{a+b+c}\) is indeed less than 1. The roots are 1 and a value less than 1. So, one of the roots is always less than 1.

Statement 1 is correct.

Evaluating Statement 2: Roots when a, b, c are Negative

Statement 2 says: One of the roots of the equation is always negative if \(a\), \(b\), and \(c\) are all negative.

The roots are \(x_1=1\) and \(x_2 = \frac{a+b-c}{a+b+c}\). Since \(x_1=1\) is positive, this statement implies that the other root, \(x_2\), must be negative when \(a, b, c < 0\).

Let's assume \(a < 0\), \(b < 0\), and \(c < 0\). We need to check if \(x_2 < 0\).

\(\frac{a+b-c}{a+b+c} < 0\)

Since \(a, b, c\) are all negative, their sum \(a+b+c\) is also negative. When multiplying or dividing an inequality by a negative number, we must reverse the direction of the inequality sign.

\(a+b-c > 0 \cdot (a+b+c)\)

\(a+b-c > 0\)

This inequality \(a+b-c > 0\) must always be true for \(x_2\) to be negative when \(a, b, c < 0\).

Let's test with some distinct negative values for \(a\), \(b\), and \(c\).

Example 1: Let \(a=-1\), \(b=-2\), \(c=-3\). These are distinct negative numbers.

Check if \(a+b-c > 0\): \((-1) + (-2) - (-3) = -1 - 2 + 3 = 0\). Since \(0\) is not greater than \(0\), the condition \(a+b-c > 0\) is not met. In this case, \(x_2 = \frac{a+b-c}{a+b+c} = \frac{0}{-1-2-3} = \frac{0}{-6} = 0\). The root is 0, which is not negative.

Example 2: Let \(a=-3\), \(b=-2\), \(c=-1\). These are distinct negative numbers.

Check if \(a+b-c > 0\): \((-3) + (-2) - (-1) = -3 - 2 + 1 = -4\). Since \(-4\) is not greater than \(0\), the condition \(a+b-c > 0\) is not met. In this case, \(x_2 = \frac{a+b-c}{a+b+c} = \frac{-4}{-3-2-1} = \frac{-4}{-6} = \frac{2}{3}\). The root is positive.

Since we found examples where \(a, b, c\) are negative but the second root \(x_2\) is not negative (it can be zero or positive), Statement 2 is not always true.

Statement 2 is incorrect.

Conclusion

Based on our analysis:

• Statement 1 is correct because if \(a, b, c > 0\), one root is 1 and the other root is \(\frac{a+b-c}{a+b+c}\), which is always less than 1.
• Statement 2 is incorrect because if \(a, b, c < 0\), one root is 1 and the other root \(\frac{a+b-c}{a+b+c}\) is not always negative; it can be zero or positive.

Therefore, only Statement 1 is correct.

Revision Table: Key Concepts for Quadratic Equations

When studying quadratic equations like \(\small Ax^2 + Bx + C = 0\), understanding the relationship between coefficients and roots is crucial.

• Roots: The values of \(x\) that satisfy the equation. A quadratic equation has two roots (which can be real or complex, distinct or repeated).
• Quadratic Formula: The roots are given by \(x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}\).
• Discriminant (\(\small \Delta\)): \(\small \Delta = B^2 - 4AC\). It determines the nature of the roots:
  • \(\small \Delta > 0\): Two distinct real roots.
  • \(\small \Delta = 0\): One real root (repeated root).
  • \(\small \Delta < 0\): Two distinct complex roots.
• Sum of Roots: \(\small x_1 + x_2 = -\frac{B}{A}\).
• Product of Roots: \(\small x_1 x_2 = \frac{C}{A}\).
• Coefficient Signs: The signs of coefficients \(A\), \(B\), and \(C\) influence the location and nature of the roots. For example, if \(C/A\) is negative, the roots must have opposite signs (one positive, one negative).

Additional Information: Properties of Quadratic Roots Analysis

Analyzing the roots of a quadratic equation often involves more than just finding their values. We can use properties like the sum and product of roots, or analyze the sign of the discriminant, to answer questions about the roots without calculating them directly.

In this problem, recognizing that \(x=1\) is a root was a key step. This is a useful technique when the coefficients involve parameters; checking simple integer values like \(x=0, 1, -1\) can sometimes reveal a root quickly.

Once one root is known, finding the other is simplified using the sum or product of roots formulas. This method is often faster than using the full quadratic formula, especially in competitive exams.

The conditions on \(a, b, c\) being real and distinct are important. Distinctness ensures \(a, b, c\) are not equal, preventing cases like \(a=b=c=0\) (which makes the equation trivial) or \(a=b\) while \(c\) is different (which doesn't change the structure significantly but is part of the given constraints). Being real ensures we are dealing with standard real number properties unless the discriminant leads to complex roots.

When dealing with inequalities involving parameters (like checking if \(\frac{a+b-c}{a+b+c} < 1\)), it is crucial to consider the sign of the denominator before multiplying, as it affects the inequality direction. In Statement 1, \(a+b+c > 0\), so the inequality direction is preserved. In Statement 2, \(a+b+c < 0\), so the inequality direction is reversed.