Problem 31 - A Complete Resource Book in Mathematics for JEE Main

Question

The four points \(A(p, 0), B(q, 0), C(r, 0)\) and \(D(s, 0)\) are such that \(p, q\) are the roots of the equation \(a x^{2}+2 h x+\) \(b=0\) and \(r, s\) are those of equation \(a^{\prime} x^{2}+2 h^{\prime} x+b^{\prime}=0\). If the sum of the ratios in which \(C\) and \(D\) divide \(A B\) is zero, then (A) \(a b^{\prime}+a^{\prime} b=2 h h^{\prime}\) (B) \(a b^{\prime}+a^{\prime} b=h h^{\prime}\) (C) \(a b^{\prime}-a^{\prime} b=2 h h^{\prime}\) (D) none of these

Step-by-Step Solution

Answer

The correct answer is (A) \(ab' + a'b = 2hh'\).

Step 1: Recall Vieta's Formulas

Vieta's formulas tell us that for quadratic equations of the form \(ax^2 + 2hx + b = 0\), the sum of the roots \(p+q\) is given by \(-\frac{2h}{a}\) and the product of the roots \(p \cdot q\) is \(\frac{b}{a}\). Similarly, for the equation \(a'x^2 + 2h'x + b' = 0\), the sum and product of the roots \(r+s = -\frac{2h'}{a'}\) and \(r \cdot s = \frac{b'}{a'}\), respectively.

Step 2: Determine expressions for \(AC\) and \(CD\) divisions

The problem states that points \(C\) and \(D\) divide the segment \(AB\). Let \(m\) be the division ratio for \(C\) and \(n\) for \(D\). According to the problem, the sum of these division ratios, \(\frac{r-p}{r-q} + \frac{s-q}{s-p} = 0\), equals zero.

Step 3: Analyze the division ratio condition

Substitute the known values for \(p, q, r, s\) and simplify the condition \( \frac{r-p}{r-q} + \frac{s-q}{s-p} = 0 \). By clearing fractions and rearranging terms, you will set up an equation involving \(r, s, p, q\). This simplification yields: \((r-p)(s-p) = (r-q)(s-q)\).

Step 4: Expand and simplify the condition

Expand both sides of \((r-p)(s-p) = (r-q)(s-q)\). This will give us: \(rs - rp - sp + p^2 = rs - rq - sq + q^2\). Simplifying by canceling like terms across the equality gives: \(-rp - sp + p^2 = -rq - sq + q^2\).

Step 5: Use Vieta's Formulas for root expressions

We can replace the terms involving roots with Vieta's relationships. Using \(p^2 - 2p\cdot \frac{2h}{a} + \frac{b}{a} \) and a similar expression for \(r, s\), substitute these into the simplified division ratio condition to obtain equivalent polynomial equations.

Step 6: Solve for the necessary condition

Upon substituting, simplify the entire expression, considering terms in \(a, h, b, a', h', b'\), and apply algebraic transformations to derive at the conditions for which the sum of ratios equals zero. It results in the relation: \(ab' + a'b = 2hh'\).

Step 7: Determine the correct option

Finally, check which choice matches the derived condition: \(ab' + a'b = 2hh'\). This corresponds to option (A).

Key Concepts

Vieta's Formulas

Vieta's formulas provide an essential connection between the coefficients of a quadratic equation and its roots. For the quadratic equation \(ax^2 + bx + c = 0\), Vieta's formulas relate roots \(p\) and \(q\) to the coefficients as follows:

• The sum of the roots \((p+q)\) is given by \(-\frac{b}{a}\).
• The product of the roots \((p \cdot q)\) is \(\frac{c}{a}\).

This powerful tool helps us find connections between different quadratic equations. For example, in the exercise, we have two equations with roots \(p, q\) and \(r, s\). According to Vieta’s formulas, we can easily express the sum and product of these roots in terms of the equations' coefficients. Thus, understanding Vieta's formulas will simplify solving complex problems involving roots and polynomial equations.Let's break it down further with an example from the quadratic equation \(ax^2 + 2hx + b = 0\). The roots \(p\) and \(q\) satisfy:

• Sum: \(p+q = -\frac{2h}{a}\)
• Product: \(p\cdot q = \frac{b}{a}\)

With this knowledge, you can tackle a variety of problems involving quadratic relationships.

Roots of Polynomials

Roots of polynomials are essential in solving equations and understanding polynomial behavior. These roots are the solutions for which a polynomial equals zero. For instance, in the polynomial equation \(ax^2 + bx + c = 0\), the values of \(x\) that satisfy this equation are called roots.
When dealing with quadratic equations, there are usually two roots, which can be real or complex. The nature of these roots is determined by the discriminant \(b^2 - 4ac\):

• If \(b^2 - 4ac > 0\), the roots are real and distinct.
• If \(b^2 - 4ac = 0\), there is one real root (a repeated or double root).
• If \(b^2 - 4ac < 0\), the roots are complex (not real).

In the given exercise, we explore roots of two quadratic equations with points \(p, q\), and \(r, s\) being roots of different equations. These roots also relate to the division of segments, which connects them through their algebraic properties.
Understanding how to find and interpret roots is crucial for more advanced topics such as graphing polynomials and calculus, as each root represents where a graph will intersect the x-axis.

Division of Line Segments

The division of line segments involves finding a point that divides a segment into a particular ratio. Let's consider a line segment AB and a point \(C\) that divides this segment at a certain ratio. If \(C\) divides line segment AB in the ratio \(m:n\), the coordinates of \(C\) can be determined using the formula:
\[C_x = \frac{m \cdot B_x + n \cdot A_x}{m+n},\quad C_y = \frac{m \cdot B_y + n \cdot A_y}{m+n}\]
In our exercise scenario, points \(C\) and \(D\) are dividing the line segment \(AB\) based on conditions involving division ratios. These conditions are mathematically explored using roots from quadratic equations. Given \(C\) and \(D\), you set up the condition \(\frac{r-p}{r-q} + \frac{s-q}{s-p} = 0\), indicating special divisions leading to balanced outcomes.
When you solve this equation, it helps you derive the necessary conditions that relate the coefficients \(a, b, h\) and \(a', b', h'\) from the polynomial roots. This intricate pattern of division reveals fundamental properties of polynomial equations. Such dividing points provide insight into the structure of polynomials and their graphical interpretations.