Problem 64 - A Complete Resource Book in Mathematics for JEE Main

Question

If \(\alpha, \beta\) are the roots of the equation \(x^{2}+p x+q=0\) then \(\frac{\alpha}{\beta}\) is a root of the equation (A) \(p x^{2}+\left(2 q-p^{2}\right) x+p=0\) (B) \(q x^{2}+\left(p^{2}-2 q\right) x+q=0\) (C) \(q x^{2}+\left(2 q-p^{2}\right) x+q=0\) (D) None of these

Step-by-Step Solution

Answer

Option C: \(q x^{2}+(2 q-p^{2}) x+q=0\) is correct.

Step 1: Recall Vieta's Formulas

For a quadratic equation of the form \(x^2 + px + q = 0\), Vieta's formulas state that the sum and product of the roots are given by \(\alpha + \beta = -p\) and \(\alpha \beta = q\) respectively.

Step 2: Express \(\frac{\alpha}{\beta}\) in Polynomial Equation

Let \(y = \frac{\alpha}{\beta}\). Then, \(\alpha = y \beta\). Substituting into the product of roots: \( \alpha \beta = q \Rightarrow y\beta^2 = q \Rightarrow \beta^2 = \frac{q}{y} \).

Step 3: Substitute \(\alpha = y \beta\) in Sum

The sum of the roots gives: \(\alpha + \beta = -p \Rightarrow y\beta + \beta = -p \Rightarrow \beta(y + 1) = -p \).

Step 4: Solve for \(\beta\)

From the equation \(\beta(y + 1) = -p\), find \(\beta = \frac{-p}{y + 1}\). Substitute in \(\beta^2 = \frac{q}{y}\) to get: \(\left(\frac{-p}{y+1}\right)^2 = \frac{q}{y}\).

Step 5: Simplify the Equation

Simplify \(\left(\frac{-p}{y+1}\right)^2 = \frac{q}{y}\) to get \(\frac{p^2}{(y+1)^2} = \frac{q}{y}\) or \(p^2y = q(y+1)^2\) which expands to \(p^2y = q(y^2 + 2y + 1)\).

Step 6: Rearrange into Quadratic Form

Rearrange the equation to \(q y^2 + (2q - p^2)y + q = 0\). This matches option C.

Key Concepts

Vieta's Formulas

Vieta's formulas are a set of relationships connecting the coefficients of a polynomial equation to sums and products of its roots. These formulas are particularly insightful when dealing with quadratic equations. For the quadratic equation of the form

• \(x^2 + px + q = 0\)

Vieta's formulas tell us that:

• The sum of the roots, \(\alpha + \beta\), is equal to \(-p\).
• The product of the roots, \(\alpha \beta\), is equal to \(q\).

These equations come in handy when solving for roots, as well as for simplifying expressions involving roots. By expressing relationships through these simple equations, solving complex polynomial identities becomes more intuitive. For the given problem, understanding the roles of \(p\) and \(q\) is essential for deriving the related polynomial equation.

Roots of equations

The roots of an equation are values that satisfy the equation when substituted for the variable. In polynomial equations, roots are the solutions where the polynomial evaluates to zero. For a quadratic equation like

• \(x^2 + px + q = 0\)

there are generally two roots: \(\alpha\) and \(\beta\). These roots can be real or complex numbers, depending on the value of discriminant \(b^2 - 4ac\) in the quadratic formula.In the specific exercise, \(\alpha\) and \(\beta\) are used to find another root, \(\frac{\alpha}{\beta}\), for a transformed polynomial equation. Understanding how roots relate to each other through operations and transformations is crucial in polynomial theory. It helps predict and simplify polynomial outcomes based on known root properties.

Polynomial equations

Polynomial equations are algebraic expressions with variables raised to whole number powers along with coefficients. They can often have multiple solutions, known as roots, corresponding to the degree of the polynomial.For quadratic polynomial equations, which have the general form

• \(ax^2 + bx + c = 0\)

this means there are two roots. These equations serve as the backbone for many functions and are crucial in algebra.In our exercise, the equation given

• \(x^2 + px + q = 0\)

considers the relationship between roots under transformations.The transformation involves finding a new root, \(\frac{\alpha}{\beta}\), requiring manipulation of the original polynomial using known identities. This showcases the versatility of polynomial equations and their ability to depict complex mathematical scenarios through their roots.