Problem 16 - A Complete Resource Book in Mathematics for JEE Main 2017

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 Problem

We are given a quadratic equation in the form \((a^{2}-1) x^{2}+(a-1) x+a^{2}-4 a+3=0\) which is claimed to be an identity in \(x\). This means that this equation holds true for all values of \(x\). Our task is to find the value of \(a\) that makes this statement true.

Step 2: Setting Coefficients to Zero

For the given quadratic to be an identity in \(x\), all its coefficients must be zero. Thus, we set each coefficient of \(x^2\), \(x\), and the constant term to zero. Therefore, we have the following conditions: \((1)\) \(a^2 - 1 = 0\), \((2)\) \(a - 1 = 0\), and \((3)\) \(a^2 - 4a + 3 = 0\).

Step 3: Solving First Condition

Solve the equation \(a^2 - 1 = 0\). This is a difference of squares, so it factors as \((a - 1)(a + 1) = 0\). Therefore, \(a = 1\) or \(a = -1\).

Step 4: Solving Second Condition

Solve \(a - 1 = 0\), which gives \(a = 1\). This condition restricts \(a\) to be exactly 1.

Step 5: Checking Consistency with Third Condition

Now, check if \(a = 1\) satisfies the third condition \(a^2 - 4a + 3 = 0\). Substitute \(a = 1\) into the equation: \(1^2 - 4 \cdot 1 + 3 = 1 - 4 + 3 = 0\), which holds true. Thus, \(a = 1\) is consistent with all three conditions.

Step 6: Conclusion

Therefore, the value of \(a\) that makes the given equation an identity in \(x\) is \(a = 1\).

Key Concepts

Quadratic Equations

Quadratic equations are a fundamental component of algebra and appear frequently in various mathematical contexts. A quadratic equation is typically in the form of \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants known as coefficients. The highest power of \( x \) in a quadratic equation is 2, indicating it is a second-degree polynomial. Understanding how to solve these equations is essential for comprehending more complex mathematical topics.To find the solution(s) of a quadratic equation, we aim to determine the value(s) of \( x \) that satisfy the equation. The standard methods include factoring, completing the square, and using the quadratic formula given by:

• \( x = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \)

Here's a key point: a quadratic equation can have one, two, or no real solutions depending on the discriminant \( b^2 - 4ac \). If the discriminant is:

• Positive, there are two distinct real roots.
• Zero, there is exactly one real root (also known as a repeated root).
• Negative, there are no real roots, but two complex roots.

In the context of the exercise, the problem presents a quadratic identity which must hold true for all \( x \), requiring an analysis of coefficients to ensure they equate to zero.

Difference of Squares

The difference of squares is a special factorization technique used in algebra. It involves expressions that can be written in the form \( a^2 - b^2 \), which can be factored as \( (a - b)(a + b) \). Recognizing a difference of squares is crucial for simplifying expressions and solving equations efficiently.In the given exercise, solving \( a^2 - 1 = 0 \) utilizes this property. This expression can be rephrased as a difference of squares, \( a^2 - 1^2 \), making it easier to factor:

• \( (a - 1)(a + 1) = 0 \)

This provides the solutions \( a = 1 \) or \( a = -1 \). Using the difference of squares enables us to break down the expression into simpler parts and quickly identify solutions, which is a powerful tool in solving algebraic equations. Understanding the formula is integral for analyzing and working with polynomials in advanced mathematics.

Coefficients Zero Method

When faced with an equation that is claimed to be an identity, such as the one in the exercise \( \left(a^2-1\right) x^2+(a-1) x+a^2-4a+3=0 \), the 'Coefficients Zero Method' becomes a critical approach. An identity must hold true for all values of the variable, which, in this case, is \( x \). To ensure this condition is met, each coefficient of the polynomial must be zero.For the exercise, the equation was split into separate conditions based on setting each coefficient to zero:

• \( a^2 - 1 = 0 \)
• \( a - 1 = 0 \)
• \( a^2 - 4a + 3 = 0 \)

By solving each of these conditions, we find the permissible values of \( a \). Each represents a constraint that must be met for the equation's polynomial identity to be valid. These conditions help verify that a singular value of \( a \), namely \( a = 1 \), satisfies all conditions, thereby confirming the given equation is indeed an identity. The method demonstrates a systematic approach to solving higher-order equations and ensuring their validity unconditionally.