Problem 85 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\left(a x^{2}+b x+c\right) y+a^{\prime} x^{2}+b^{\prime} x+c^{\prime}=0\), then the condi- tion that \(x\) may be a rational function of \(y\) is (A) \(\left(a c^{\prime}-a^{\prime} c\right)^{2}=\left(a b^{\prime}-a^{\prime} b\right)\left(b c^{\prime}-b^{\prime} c\right)\) (B) \(\left(a b^{\prime}-a^{\prime} b\right)^{2}=\left(a c^{\prime}-a^{\prime} c\right)\left(b c^{\prime}-b^{\prime} c\right)\) (C) \(\left(b c^{\prime}-b^{\prime} c\right)^{2}=\left(a b^{\prime}-a^{\prime} b\right)\left(a c^{\prime}-a^{\prime} c\right)\) (D) None of these

Step-by-Step Solution

Answer

The correct answer is option (A).

Step 1: Rewrite the Given Equation

We start with the given expression \((ax^2 + bx + c)y + a'x^2 + b'x + c' = 0\). Let's rearrange it: \( ax^2y + bxy + cy + a'x^2 + b'x + c' = 0 \). Here, our aim is to find conditions where \(x\) can be a rational function of \(y\).

Step 2: Group Terms by Powers of x

Rewrite the equation by grouping similar powers of \(x\): \((ay + a')x^2 + (by + b')x + cy + c' = 0\). This is now a quadratic in terms of \(x\).

Step 3: Apply the Condition for Quadratic in x to Have Rational Roots

For \(x\) to be a rational function of \(y\), the discriminant of this quadratic equation, \((by + b')^2 - 4(ay + a')(cy + c')\), must be zero (perfect square).

Step 4: Simplify the Expression

Simplify the discriminant: \((by + b')^2 - 4achy^2 - 4(accy + b'cy + bc'y + b'c') = 0\). Group terms, if necessary, to correlate it to one of the given specific forms.

Step 5: Equate and Identify the Condition

The simplification leads us to compare the discriminant with the form \((a c' - a' c)^2 = (a b' - a' b)(b c' - b' c)\) which matches option (A).

Step 6: Conclude the Correct Option

The condition derived meets option (A): \((a c' - a' c)^2 = (a b' - a' b)(b c' - b' c)\). Thus, this is the correct choice when \(x\) can be expressed as a rational function of \(y\).

Key Concepts

Quadratic Equations

At the heart of many algebra problems, you'll find quadratic equations. These are equations written in the form \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants, and \(x\) represents an unknown variable. In this exercise, you are dealing with a special type of quadratic equation where the coefficients themselves depend on another variable \(y\). This equation becomes \((ay + a')x^2 + (by + b')x + cy + c' = 0\).

Key characteristics of quadratic equations include their parabolic graphs and their solutions, which are the \(x\)-values where the graph intersects the \(x\)-axis. These solutions can be found using methods like factoring, completing the square, or the quadratic formula. For solutions to be useful in finding rational functions, as in this exercise, the roots should be rational, which leads us to explore the discriminant as a determining factor.

Discriminant

The discriminant is a special expression that helps us understand the nature of the roots of a quadratic equation without necessarily solving it. For a standard quadratic equation \(ax^2 + bx + c = 0\), the discriminant \(D\) is given by the formula \(D = b^2 - 4ac\).

In this exercise, we look at a more complex form: the discriminant for the equation \((ay + a')x^2 + (by + b')x + cy + c' = 0\). Here, the discriminant is \((by + b')^2 - 4(ay + a')(cy + c')\). The value of this discriminant tells us:

• If \(D > 0\), there are two distinct real roots.
• If \(D = 0\), there is exactly one real root, which is a rational double root (perfect square), the condition needed for \(x\) to be a rational function of \(y\).
• If \(D < 0\), the roots are not real numbers but complex conjugates.

In this problem, the discriminant condition simplifies to show that \((a c' - a' c)^2 = (a b' - a' b)(b c' - b' c)\), indicating rational roots.

Rational Roots

Rational roots are specific solutions to quadratic equations that can be expressed as a ratio of two integers, which makes them particularly neat and practical to work with in calculations. When we talk about rational roots in quadratic equations, we're interested in the circumstances under which these roots can be achieved.

Using the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), the roots are rational if and only if the discriminant \(\sqrt{b^2 - 4ac}\) is a perfect square. In other words, there must be an integer \(k\) such that \(b^2 - 4ac = k^2\).

In the problem at hand, this concept is applied to determine when \(x\), a supposed rational function of \(y\), fits this criteria under the conditions set by the discriminant. Thus, by manipulating the equation into the correct form and analyzing the discriminant for perfect square conditions, we determined that option (A) holds true: \((a c' - a' c)^2 = (a b' - a' b)(b c' - b' c)\). This ensures that \(x\) indeed can be expressed rationally in terms of \(y\).