Problem 16 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\left(a^{2}-1\right) x^{2}+(a-1) x+a^{2}-4 a+3=0\) is an identity in \(x\), then the value of \(a\) is (A) 1 (B) 3 (C) \(-1\) (D) \(-3\)

Step-by-Step Solution

Answer

The value of \(a\) is 1.

Step 1: Understanding the Identity

An identity in \(x\) means that the equation is true for all values of \(x\). This implies that the coefficients of \(x^2\), \(x\), and the constant term are all zero.

Step 2: Equating the Coefficient of \(x^2\) to Zero

The coefficient of \(x^2\) is \(a^2 - 1\). For the equation to be an identity, \(a^2 - 1 = 0\).

Step 3: Solving \(a^2 - 1 = 0\)

The equation \(a^2 - 1 = 0\) can be factored into \((a - 1)(a + 1) = 0\). Solving this gives \(a = 1\) or \(a = -1\).

Step 4: Equating the Coefficient of \(x\) to Zero

The coefficient of \(x\) is \(a - 1\). Setting \(a - 1 = 0\) gives \(a = 1\).

Step 5: Confirming with the Constant Term

Substitute \(a = 1\) into the constant term \(a^2 - 4a + 3\), which becomes \(1^2 - 4 \times 1 + 3 = 0\). Therefore, \(a = 1\) satisfies the identity.

Step 6: Verify if \(a = -1\) Satisfies the Constant Term

For \(a = -1\), substitute into the constant term: \((-1)^2 - 4(-1) + 3 = 1 + 4 + 3 = 8 eq 0\). Therefore, \(a = -1\) does not satisfy the identity.

Key Concepts

Coefficient Comparison

When working with polynomial identities, one fundamental approach is 'Coefficient Comparison.' This technique involves setting each coefficient of the polynomial to zero when the polynomial is equal to zero. This means that to verify if an equation is an identity, each term's coefficient must equal zero.

In the context of the given polynomial identity \(\left(a^{2}-1\right) x^{2}+(a-1) x+a^{2}-4 a+3=0\), coefficient comparison is key. For the equation to be true for all values of \(x\), which defines it as an identity, all coefficients must vanish:

• The coefficient of \(x^2\) is \(a^2 - 1\), and setting this to zero gives \(a^2 - 1 = 0\).
• For \(x\), the coefficient is \(a - 1\), leading to \(a - 1 = 0\).
• The constant term also emerges as part of the equation's complete set of conditions.

This approach allows us to solve systematically for the variable \(a\) by comparing each coefficient and setting them to zero.

Quadratic Equations

Quadratic equations often arise when dealing with polynomials, typically in the form \(ax^2 + bx + c = 0\). In this exercise, the quadratic equation features prominently, requiring us to find values for \(a\) that satisfy certain conditions.

The coefficient of \(x^2\) in the original problem, \(a^2 - 1 = 0\), is a simplified form of a quadratic equation. Solving this particular equation involves:

• Rewriting it as \(a^2 - 1 = (a - 1)(a + 1) = 0\).
• The solutions to this equation are the roots \(a = 1\) and \(a = -1\).

Solving quadratic equations through factoring, as demonstrated, is an essential skill in algebra. The defined roots reveal potential values of \(a\) that could satisfy an identity across all terms of the polynomial.

Factoring

Factoring is a method used to simplify polynomials by rewriting them as products of simpler polynomials. This process helps in solving equations as it can easily reveal the roots or solutions of a polynomial.

In the given exercise, factoring is crucial twice:

• First, in Step 3, we factor \(a^2 - 1 = (a - 1)(a + 1)\). This reveals the potential values of \(a\), showing the targets to ensure zeroing-out the coefficient in the equation.
• Next, when checking if \(a = 1\) satisfies the equation completely, we substitute back and explore if it validates all terms becoming zero. It confirms \(a = 1\) is a solution as \(1^2 - 4(1) + 3 = 0\).

Factoring leads to simplified equations that are easier to solve, providing a clear path to identifying polynomial identities.