Problem 106 - A Complete Resource Book in Mathematics for JEE Main

Question

The quadratic equation \(\frac{(x+b)(x+c)}{(b-a)(c-a)}+\frac{(x+c)(x+a)}{(c-b)(a-b)}+\frac{(x+a)(x+b)}{(a-c)(b-c)}=1\) has (A) two real and distinct roots (B) imaginary roots (C) equal roots (D) infinite roots

Step-by-Step Solution

Answer

The solution involves exploiting symmetry to find the answer.

Step 1: Simplify the Equation

The given quadratic equation is \( \frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} = 1 \). First, we can note that this equation is symmetric in cyclic order of the parameters \(a, b, c\). In such cases, we explore the equation by assuming symmetry properties or substitutions.

Step 2: Substitute for Simplicity

Using symmetry properties or checking for a symmetry case can help simplify. If we set \(x = a\), the terms \((x + b)(x + c)\), \((x + c)(x + a)\), and \((x + a)(x + b)\) become clearly zero or can cancel out.

Key Concepts

Real and Imaginary Roots

When studying quadratic equations, understanding the nature of the roots is crucial. Quadratic equations can have roots that are:

• Real and distinct - where the equation intersects the x-axis at two different points
• Real and equal - when the parabola touches the x-axis at a single point, known as the vertex
• Complex or imaginary - if there are no intersections with the x-axis, indicating complex solutions

For instance, consider the quadratic equation in standard form: \[ ax^2 + bx + c = 0 \] The roots can be described using the discriminant, \(b^2 - 4ac\): - **If positive**: the equation has two real and distinct roots. - **If zero**: the equation has one real root (a repeated root). - **If negative**: the equation has two complex roots. These concepts help us systematically predict the type of roots without needing to solve the equation each time.

Symmetrical Equations

Symmetrical equations exhibit a specific property where the equation remains unchanged under certain transformations. This symmetry can simplify the problem-solving process. In the case of our quadratic equation: \[ \frac{(x+b)(x+c)}{(b-a)(c-a)} + \frac{(x+c)(x+a)}{(c-b)(a-b)} + \frac{(x+a)(x+b)}{(a-c)(b-c)} = 1 \] The equation is symmetric in the parameters \(a, b, c\). This means swapping or permuting these parameters does not change the form of the equation. Why is symmetry useful?

• It allows simplifications by reducing the number of terms or entire factors.
• It enables the use of substitution methods, where variables can be set equal to each other based on symmetry, making it easier to solve.

In our equation, recognizing symmetry led us to explore setting \(x = a\), simplifying each term and potentially cancelling complex parts without solving fully. This property is often used in algebra to understand and solve more intricate equations efficiently.

Substitution Method

The substitution method is a powerful technique used to solve equations, particularly useful in algebra. It involves substituting one variable with another expression, simplifying the equation, and making it easier to solve. In our quadratic equation, we use substitution to capitalize on the symmetry noticed earlier. By setting \(x = a\):

• The equation can be simplified as terms that involve \((x + b)(x + c)\), \((x + c)(x + a)\), and \((x + a)(x + b)\) become easier to handle.
• This reduces the equation to a form where terms either cancel out or transform into zero. This simplification can lead directly to recognizing potential roots without further calculation.

The substitution method is particularly helpful when dealing with complex equations that possess underlying patterns or symmetries. By strategically choosing which substitutions to make, we can break down even intricate equations into more manageable parts.