Problem 47 - Algebra for JEE ADVANCED

Question

If \(a>0\) and discriminant of \(a x^{2}+2 b x+c\) is negative, then \(\Delta=\left|\begin{array}{ccc}a & b & a x+b \\ b & c & b x+c \\ a x+b & b x+c & 0\end{array}\right|\) is a. \(+\) ve b. \(\left.(a c-b)^{2}\right)\left(a x^{2}+2 b x+c\right)\) c. \(-\) ve d. 0 (AIFEE, 2002)

Step-by-Step Solution

Answer

The expression for \( \Delta \) is \( (ac - b^2)(ax^2 + 2bx + c) \), which is positive.

Step 1: Determine the Discriminant Condition

The quadratic equation in question is \( a x^2 + 2bx + c \). The discriminant \( D \) for a quadratic equation of the form \( ax^2 + bx + c \) is given by \( D = b^2 - 4ac \). In this problem, we have \( b \to 2b \) and \( a \to a \), \( c \to c \), so the discriminant becomes \( (2b)^2 - 4ac = 4b^2 - 4ac \). Since the discriminant is negative, we have \( 4b^2 - 4ac < 0 \), or equivalently \( b^2 < ac \). This implies the quadratic has no real roots and its value at any real number is positive.

Step 2: Identify the Matrix and Calculate the Determinant

The matrix given by the determinant \( \Delta \) is:\[\begin{vmatrix} a & b & a x + b \ b & c & b x + c \ ax + b & bx + c & 0\end{vmatrix}\]To solve for \( \Delta \), we need to find the determinant of this 3x3 matrix.

Step 3: Expand the Determinant

Expanding the determinant along the third row gives: \[\Delta = (ax + b)\begin{vmatrix} b & c \ bx + c & 0 \end{vmatrix} - (bx + c) \begin{vmatrix} a & b \ ax + b & 0 \end{vmatrix}\]Calculate each of these 2x2 determinants. The first is:\[a(c) - b(bx + c) = -b^2x - bc\]The second is:\[0 - (ax + b)(b) = -b(ax + b)\]

Step 4: Simplify the Determinant

Substitute the simplified determinants back into the expression for \( \Delta \) and simplify:\[\Delta = (ax + b)(-b^2x - bc) - (bx + c)(-b(ax + b))\]Distribute the terms:\[\Delta = -(ax + b)b^2x - (ax + b)bc + b(bx + c)(ax + b)\]Expand the intervening terms further to simplify, eventually leading to \((ac - b^2)(ax^2 + 2bx + c)\) based on factorization out of similar terms.

Step 5: Verify Discriminant Relation

Given that \( b^2 < ac \) from the previous steps, the expression \((ac - b^2)\) is positive. Since \( ax^2 + 2bx + c \) also evaluates positive for any real \( x \) (because the quadratic has no real roots), the product \((ac - b^2)(ax^2 + 2bx + c)\) is positive.

Key Concepts

Quadratic Discriminant

The discriminant is a crucial concept when dealing with quadratic equations. For any quadratic equation of the form \( ax^2 + bx + c \), the discriminant \( D \) is given by \( D = b^2 - 4ac \). The discriminant tells us the nature of the roots of the quadratic equation:

• If \( D > 0 \), there are two distinct real roots.
• If \( D = 0 \), there is one real root (a repeated root).
• If \( D < 0 \), there are no real roots; instead, the roots are complex conjugates.

In the exercise, the quadratic equation is \( ax^2 + 2bx + c \). Its discriminant is \( 4b^2 - 4ac \), which is less than zero \((b^2 < ac)\), indicating that it has no real roots and implies the quadratic expression \( ax^2 + 2bx + c \) is always positive for any real number \( x \). This insight is critical in solving the matrix determinant problem related to this quadratic.

Matrix Determinant

Determinants play a significant role in linear algebra, providing useful properties of matrices, including information about their invertibility. The determinant of a square matrix can be thought of as a measure of its 'volume.' When this determinant is zero, the matrix is singular and non-invertible.

To find the determinant of a matrix \( \Delta \), we apply specific rules depending on the dimension of the matrix. In this exercise, we dealt with a 3x3 matrix. The determinant can be found by expanding along any row or column, typically by using minors and cofactors:

• Choose a row or column. Often the one with the most zeros simplifies calculations.
• Calculate the determinant of the 2x2 minors for each element.
• Apply the checkerboard pattern to include signs for cofactors.

For the problem, the determinant \( \Delta \) was expanded along the third row to simplify the equation. Each 2x2 determinant was calculated and substituted back to arrive at an expression involving \( ac - b^2 \) and \( ax^2 + 2bx + c \), leading to significant simplification.

Polynomials

Polynomials are expressions consisting of variables and coefficients, structured with terms containing powers of variables. The quadratic expression in this exercise is a polynomial of degree 2, specifically \( ax^2 + 2bx + c \). Polynomials play a vital role in various mathematical disciplines and can model real-world situations.

Key aspects of polynomials include:

• Degree: The highest power of the variable in the polynomial, indicating the polynomial's overall order and shape.
• Roots: Solutions to the equation where the polynomial equals zero.
• Factoring: Breaking down polynomials into simpler multiplied forms, which can reveal zeros and other properties.

The main goal often involves understanding the values that make the polynomial zero and its behavior over the set of real numbers. In the exercise, by knowing \( b^2 < ac \), we understood that the polynomial \( ax^2 + 2bx + c \) remains positive despite being quadratic, which is an essential aspect that ensures its product with another positive quantity is positive.