Problem 116 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(c \neq 0\) and the equation \(\frac{p}{2 x}=\frac{a}{x+c}+\frac{b}{x-c}\) has two equal roots, then \(p\) can be (A) \((\sqrt{a}-\sqrt{b})^{2}\) (B) \((\sqrt{a}+\sqrt{b})^{2}\) (C) \(a+b\) (D) \(a-b\)

Step-by-Step Solution

Answer

Option (A) \\((\sqrt{a}-\sqrt{b})^{2}\\) is the correct answer.

Step 1: Understand Two Equal Roots

When an equation has two equal roots, its discriminant equals zero. This is our initial condition that will determine how we handle the equation's terms to find \(p\).

Step 2: Set Up the Equation

Start from the given equation: \( \frac{p}{2x} = \frac{a}{x+c} + \frac{b}{x-c} \). To proceed, we should find a common denominator for the terms on the right-hand side.

Step 3: Clear the Denominators

Multiply the entire equation by \((x+c)(x-c)\) to eliminate the denominators, yielding: \( p(x+c)(x-c) = 2x(a(x-c) + b(x+c)) \). Simplify to obtain a quadratic equation.

Step 4: Simplify and Form Quadratic Equation

Simplify both sides: \( p(x^2 - c^2) = 2x(ax - ac + bx + bc) \). This simplifies to \( px^2 - pc^2 = 2(ax^2 - acx + bx^2 + bcx) \), which becomes \( px^2 - 2(ax^2 + bx^2) = 2x(bc - ac) \).

Step 5: Combine Like Terms

Combine like terms to form \((p - 2a - 2b)x^2 = 2c(b - a)x \). Now we have a quadratic equation in standard form \(Ax^2 + Bx + C = 0\), with coefficients \(A = p - 2a - 2b\), \(B = 2c(b - a)\), and \(C = 0\).

Step 6: Apply Discriminant Condition

For two equal roots, the discriminant \( B^2 - 4AC = 0 \). With \(A = p - 2a - 2b\), \(B = 2c(b-a)\), and \(C = 0\), the discriminant simplifies to \((2c(b-a))^2 = 0\).

Step 7: Solve for Condition on \(p\)

Since the discriminant is zero when \((2c(b-a)) = 0\), it implies \(b = a\). Substitute \(b = a\) into \(p = (\sqrt{a} - \sqrt{b})^2\), which simplifies to \(p = 0\). Check options to determine that \((\sqrt{a} - \sqrt{b})^2 = 0\) is valid when \(a = b\), which corresponds to option (A).

Key Concepts

Discriminant

The discriminant is a crucial component in quadratic equations, helping to determine the nature of the roots. For a quadratic equation of the form \(Ax^2 + Bx + C = 0\), the discriminant is given by the formula \(B^2 - 4AC\). The discriminant provides insight into the number and type of roots the equation has:

• If the discriminant is greater than zero, the quadratic equation has two distinct real roots.
• If it equals zero, the equation has two equal roots, or one real repeated root.
• If the discriminant is less than zero, the equation has no real roots, but instead, two complex roots.

When the exercise states that the equation has two equal roots, it means the discriminant must be zero. This aligns the problem toward a specific type of solution, which then impacts the calculations involving the variables, such as \(p\) in this case.

Equal Roots

Equal roots in a quadratic equation occur when both solutions for \( x \) are the same. Mathematically, this is expressed by the condition that the discriminant \(B^2 - 4AC\) is zero. This condition simplifies the problem since it means the quadratic will touch the x-axis at exactly one point.

The equation from the exercise transforms into a standard quadratic form, allowing us to apply the discriminant condition:

• The coefficients \(A\), \(B\), and \(C\) are extracted from the rewritten equation.
• Setting the discriminant to zero gives the condition that needs to be met for the roots to be equal.

In this problem, it shows that when \(b = a\), the discriminant becomes zero, confirming the equal roots for the quadratic equation.

Rational Expressions

Rational expressions are fractions that involve polynomials in their numerators and/or denominators. They offer a structured way to handle non-linear equations but can be notoriously complex when solving quadratic equations like the one given in the exercise.

In the given exercise, rewriting the equation into rational expressions \(\frac{a}{x+c}\) and \(\frac{b}{x-c}\) helps understand each part of the equation separately:

• The common denominator is obtained through multiplying the denominators \((x+c)(x-c)\).
• Elimination of the denominators by multiplying both sides results in a cleaner, solvable equation.

Simplifying these rational expressions by clearing the denominators allows us to identify a quadratic equation, which is crucial for applying discriminant tests to find if \(p\) could take certain values based on the roots provided.