Problem 86 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

If \(\alpha, \beta\) are the roots of \(x^{2}+p x+q=0\) and \(\gamma, \delta\) are the roots of \(x^{2}+r x+s=0\), evaluate \((\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)\) in terms of \(p, q, r\) and \(s\). Hence deduce the condition that the equations have a common root.

Step-by-Step Solution

Answer

The expression \((\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)\) in terms of p, q, r, and s can be written as \((p^2+r^2)^2\). For the two given quadratic equations to have a common root, their coefficients of the linear term must be equal or opposite in signs (p = r or p = -r).

Step 1: 1. Finding the sum and product of roots

By using Vieta's formulas, we find the sum and product of roots for the given quadratic equations: For the equation \(x^2+px+q=0\), the sum of roots is \(\alpha+\beta=-p\), and the product of roots is \(\alpha\beta=q\). For the equation \(x^2+rx+s=0\), the sum of roots is \(\gamma+\delta=-r\), and the product of roots is \(\gamma\delta=s\).

Step 2: 2. Expanding the expression

Now, our task is to expand and simplify the given expression: \((\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)\) Expanding it step by step, we get: \([(\alpha-\gamma)(\alpha-\delta)][(\beta-\gamma)(\beta-\delta)]\) \([(\alpha^2-\alpha\delta-\alpha\gamma+\gamma\delta)][(\beta^2 - \beta\delta-\beta\gamma+\gamma\delta)]\)

Step 3: 3. Substituting the sum and product of roots

Now, substitute the values obtained in step 1: \(\alpha+\beta=-p, \alpha\beta=q, \gamma+\delta=-r, \gamma\delta=s\) We need to find the values of \(\alpha\gamma, \alpha\delta, \beta\gamma\) and \(\beta\delta\) in terms of p, q, r, s. \(\alpha\gamma + \alpha\delta = \alpha(\gamma + \delta) \implies \alpha(-r)\) \(\beta\gamma + \beta\delta = \beta(\gamma + \delta) \implies \beta(-r)\) \(\alpha\gamma\beta + \alpha\beta\delta = (\alpha\beta)(\gamma +\delta) \implies q(-r)\) Now, substitute these values into the expanded expression: \([(\alpha^2-\alpha\delta-\alpha\gamma+\gamma\delta)][(\beta^2 - \beta\delta-\beta\gamma+\gamma\delta)]\) \([((\alpha+\beta)^2+(\gamma+\delta)^2)][((\alpha+\beta)^2+(\gamma+\delta)^2)]\) \([(p^2+r^2)][(p^2+r^2)]\)

Step 4: 4. Simplifying the expression

We now simplify the expression: \((p^2+r^2)^2\) So, the expression \((\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)\) can be written as \((p^2+r^2)^2\) in terms of p, q, r, and s.

Step 5: 5. Finding the condition for a common root

For the two equations to have a common root, the expression \((\alpha-\gamma)(\alpha-\delta)(\beta-\gamma)(\beta-\delta)\) must be equal to zero: \((p^2+r^2)^2=0\) \(p^2+r^2=0\) \(p^2=-r^2\) This implies that p = \(\pm r\). So, the condition for the two given quadratic equations to have a common root is either their coefficients of the linear term are equal or opposite in signs (p = r or p = -r).

Key Concepts

Vieta's Formulas

Understanding Vieta's formulas is essential for working with polynomial equations, particularly quadratics. These formulas link the coefficients of a polynomial to sums and products of its roots. Specifically, for a quadratic equation of the form \(ax^{2}+bx+c=0\), with roots \(\rho_1\) and \(\rho_2\), Vieta's formulas state that \(\rho_1 + \rho_2 = -\frac{b}{a}\) and \(\rho_1 \cdot \rho_2 = \frac{c}{a}\).

These relationships are derived from the fact that any polynomial can be factored as \(a(x - \rho_1)(x - \rho_2)\), and expanding this product will give the original polynomial, with the coefficients relating directly to the roots. For our quadratic equations, knowing the sum and product of roots provides a straightforward way to understand their properties and relationships, such as when exploring the condition for common roots between two quadratics.

Sum and Product of Quadratic Roots

The sum and product of roots of a quadratic equation offer vital information about the equation itself. When the quadratic equation is given by \(x^{2} + px + q = 0\), the sum of the roots \(\rho_1 + \rho_2\) equals \(-p\), and the product \(\rho_1 \rho_2\) equals \(q\), according to Vieta’s formulas.

This concept proves extremely useful in many areas of algebra, including simplifying expressions and solving problems without explicitly finding the roots. For instance, in the given example, to find the value of \((\rho_1-\rho_3)(\rho_1-\rho_4)(\rho_2-\rho_3)(\rho_2-\rho_4)\), we can manipulate these sums and products to express complex expressions in terms of the coefficients of the equations.

Symmetric Functions of Roots

Symmetric functions of roots are expressions constructed from the roots of a polynomial that remain unchanged under any permutation of the roots. For quadratic equations, functions like \(\rho_1 + \rho_2\) and \(\rho_1 \rho_2\) are symmetric because swapping \(\rho_1\) and \(\rho_2\) leaves these expressions the same.

This property of symmetry facilitates the ability to relate these expressions directly to the coefficients of the polynomial, rendering tasks such as finding a specific function of the roots manageable without having to solve for the roots directly. Symmetric functions are particularly prominent when solving systems of equations or when evaluating complex expressions involving roots, as was the case in the earlier exercise.

Conditions for Common Roots

Determining conditions for common roots in quadratic equations is a frequent exercise in algebra. The common root between two quadratics exists where there is at least one root that they share. We use specific criteria stemming from the relationships between the coefficients of the equations, as noted by the expressions derived from Vieta’s formulas and symmetric functions.

For instance, in the provided exercise, we arrive at a necessary condition for a common root by equating to zero the expression we evaluated in terms of the coefficients. This led to the realization that for a common root to exist between the two given quadratic equations, the coefficients of their linear terms must be equal or opposite in sign, \(p = \pm r\). This logical deduction is critical for students to understand how to determine the link between two equations having a shared solution.