Problem 95 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(q \neq 0\) and the equation \(x^{3}+p x^{2}+q=0\) has a root of multiplicity 2 , then \(p\) and \(q\) are connected by (A) \(p^{2}+2 q=0\) (B) \(p^{2}-2 q=0\) (C) \(4 p^{3}+27 q+1=0\) (D) \(4 p^{3}+27 q=0\)

Step-by-Step Solution

Answer

The correct option is (A) \(p^{2}+2q=0\).

Step 1: Understanding Root Multiplicity

If the polynomial \(x^{3}+p x^{2}+q=0\) has a root \(\alpha\) of multiplicity 2, it means \((x-\alpha)^{2}\) is a factor of the polynomial.

Step 2: Express the Polynomial with Known Roots

Since \(\alpha\) is a root of multiplicity 2, the polynomial can be expressed as \((x-\alpha)^2(x-\beta)=0\). Expanding gives us \((x^2 - 2\alpha x + \alpha^2)(x - \beta) = x^3 - (2\alpha + \beta)x^2 + (\alpha^2 + 2\alpha\beta)x - \alpha^2\beta\).

Step 3: Compare Coefficients

For \(x^3 - (2\alpha + \beta)x^2 + (\alpha^2 + 2\alpha\beta)x - \alpha^2\beta = x^3 + px^2 + q\), the coefficients of \(x^2\) and the constant term must match with the original polynomial: \(- (2\alpha + \beta) = p\) and \(-\alpha^2\beta = q\).

Step 4: Solve for \(\beta\) in Terms of \(p\) and \(\alpha\)

From the equation \(\beta = -2\alpha - p\), substitute \(\beta\) into the expression for \(q\): \[ q = -(\alpha)^2(-2\alpha - p) = 2\alpha^3 + \alpha^2p \, .\]

Step 5: Relate \(p\) and \(q\) According to the Condition

Substitute \(p = -2\alpha - \beta\) from the expression for \(q\) as \[ q = 2\alpha^3 + \alpha^2p \, ,\] into the problem condition form, we have: \[ p^2 - 2(2\alpha^3 + \alpha^2p) = 0 \, ,\]which gives \[ p^2 - 4\alpha^3 - 2\alpha^2p = 0 \, .\]

Step 6: Simplify to Find the Correct Option

Rearrange and simplify the equation:\[ p^2 + 2q = 2(\alpha^2p + 2\alpha^3) + q = 0 \, .\] As \(p^2 + 2q = 0\) aligns with option (A), p^2 + 2q = 0 is the correct answer.

Key Concepts

Understanding Root Multiplicity

When dealing with polynomial equations, the concept of root multiplicity is crucial. Multiplicity refers to the number of times a root is repeated in a polynomial equation. For instance, if we have a cubic equation, such as \(x^3 + px^2 + q = 0\), and it has a root \(\alpha\) with a multiplicity of 2, it means that \(\alpha\) is a root repeated twice. In such cases, \((x - \alpha)^2\) acts as a factor of the polynomial.

This characteristic is essential since it impacts how we can factorize and solve the polynomial. In simpler terms, if a polynomial is expressed as \((x - \alpha)^2(x - \beta) = 0\), \(\alpha\) is not just a root but a consistent repeated solution that must be considered to solve the equation thoroughly.

Identifying a root's multiplicity is key to understanding the polynomial's behavior and ensuring all solutions are accounted for in analysis and computation.

Exploring Cubic Equations

Cubic equations are polynomial equations of degree three, typically expressed in the general form \(ax^3 + bx^2 + cx + d = 0\). In these equations, \(a\), \(b\), \(c\), and \(d\) are constants with \(a eq 0\).

The presence of a cubic term \(x^3\) gives these equations their distinct nature, and they can have one, two, or three real roots. Cubic equations are particularly interesting because they often involve roots with multiplicity, such as in our problem where the equation has a double root.

In practice, solving cubic equations can be complex, requiring techniques such as factoring or using the cubic formula. Special cases, like the one in the problem where a root repeats, can sometimes be tackled through factorization, leveraging our understanding of root multiplicity and manipulating known formulas to deduce other coefficients and relationships within the equation.

Factorization of Polynomials

Factorization of polynomials is a fundamental concept that involves expressing a polynomial as a product of its simplest components or "factors." This approach can simplify solving polynomial equations, such as cubic equations, especially when these polynomials can be broken down into products involving repeated roots.

For instance, the polynomial \(x^3 + px^2 + q = 0\) with a root of multiplicity 2 can be factorized as \((x - \alpha)^2(x - \beta) = 0\). Here, understanding that \((x - \alpha)^2\) is a factor, due to the repeating root, helps in re-writing and re-analyzing the polynomial.

Factorization is not just a tool for simplifying an equation but also for revealing insights into the nature of its roots and coefficients. By comparing coefficients from the expanded form back to the original polynomial, we can identify relationships such as those between \(p\) and \(q\). This process is pivotal in tackling the exercise's problem, where the choice of the correct relationship (\(p^2 + 2q = 0\)) hinges on proper factorization and understanding of the polynomial structure.