Problem 157 - A Complete Resource Book in Mathematics for JEE Main

Question

If the equations \(x^{2}+2 x+3=0\) and \(a x^{2}+b x+c=0, a, b, c \in R\) have a common root, then \(a: b: c\) is (A) \(3: 2: 1\) (B) \(1: 3: 2\) (C) \(3: \overline{1: 2}\) (D) \(1: 2: 3\)

Step-by-Step Solution

Answer

The ratio \(a:b:c\) is \(1:2:3\), corresponding to choice (D).

Step 1: Identify Common Root

Let's denote the common root as \( \alpha \). Since \( \alpha \) is a root of the first equation \( x^2 + 2x + 3 = 0 \), it must satisfy the equation: \( \alpha^2 + 2\alpha + 3 = 0 \).

Step 2: Use Common Root in Second Equation

Since \( \alpha \) is also a root of the second equation \( a x^2 + b x + c = 0 \), it must satisfy this equation as well: \( a\alpha^2 + b\alpha + c = 0 \).

Step 3: Relate the Two Equations

From step 1, we have \( \alpha^2 + 2\alpha + 3 = 0 \). Substitute \( \alpha^2 = -2\alpha -3 \) into \( a\alpha^2 + b\alpha + c = 0 \):\[ a(-2\alpha - 3) + b\alpha + c = 0 \] This simplifies to: \[ (-2a + b)\alpha + (-3a + c) = 0 \].

Step 4: Equate Coefficients for Root Existence

Since the above equation must be true for all \( \alpha \), set the coefficients of \( \alpha \) and the constant terms to zero:1. \(-2a + b = 0 \)2. \(-3a + c = 0 \).

Step 5: Solve for Ratios of Coefficients

From the two equations:1. \(-2a + b = 0 \) gives \( b = 2a \).2. \(-3a + c = 0 \) gives \( c = 3a \).Thus, the ratio \( a : b : c \) is obtained by substituting: \( a : 2a : 3a \), which simplifies to \( 1:2:3 \).

Key Concepts

Quadratic Equations

A quadratic equation is a type of polynomial equation in which the highest power of the variable is 2. This means it takes the form of:

• ax^2 + bx + c = 0

To solve a quadratic equation, you can apply the quadratic formula, factor the equation, or complete the square. Quadratic equations are unique in that they can have as many as two roots. These roots can be real or complex depending on the discriminant, which is calculated as b² - 4ac. In the context of this exercise, we are focusing on finding a common root between two quadratic equations. This involves using one of the roots that satisfies the first equation in the second equation as well. By discovering such a root, we can establish a relationship between the coefficients of both quadratic equations.

System of Equations

A system of equations consists of two or more equations that have common solutions. In solving systems, the goal is to find values that satisfy all equations simultaneously.In this exercise, the system of equations is illustrated through the relationship between two quadratic equations:

• x² + 2x + 3 = 0
• ax² + bx + c = 0

The task is to determine how these equations relate through their shared root, denoted as \( \alpha \). When two equations have a common root, the value that satisfies one equation also satisfies the second. This provides a system where unknowns are tied together by common solutions. The process involves substituting this root into both equations and aligning terms, which allows us to explore the relationship between coefficients of the variables.

Ratio of Coefficients

The ratio of coefficients in quadratic equations helps in comparing equations to find common themes or solutions. Here, it is essential to discover the ratio of coefficients in the second equation when a common root exists with the first equation.This ratio is derived from equating the two sides of a transformed equation when substituting a common root \( \alpha \) into both equations. From the solution, the steps to find these ratios include:

• Using the common root equation \( \alpha^2 + 2\alpha + 3 = 0 \) to substitute \( \alpha^2 = -2\alpha - 3 \).
• Replacing \( \alpha^2 \) in the second equation \( a\alpha^2 + b\alpha + c = 0 \), resulting in \((-2a + b)\alpha + (-3a + c) = 0 \).
• Setting coefficients \(-2a + b = 0 \) and \(-3a + c = 0 \) to zero provides \( b = 2a \) and \( c = 3a \).

Thus, the ratio \( a : b : c \) becomes simple fractions forming \( 1:2:3 \). These coefficients depict how each term in the equation scales when they share a common root.