Problem 49 - Mathematics for IIT JEE Main and Advanced Differential Calculus Algebra Trigonometry

Question

Find the equation whose roots are \((\alpha+\beta)^{2}\) and \((\alpha-\beta)^{2}\), where \(\alpha, \beta\) are the roots of \(2 x^{2}+2(m+n) x+\left(m^{2}+n^{2}\right)=0 .\) \\{Ans. \(\left.x^{2}-4 m n x-\left(m^{2}-n^{2}\right)^{2}=0\right\\}\)

Step-by-Step Solution

Answer

The equation with the new roots is \(x^2 - 4mnx - (m^2 - n^2)^2 = 0\).

Step 1: Identify the given roots of the new equation

The roots of the new equation are given as: \( r_1 = (\alpha + \beta)^2 \) and \( r_2 = (\alpha - \beta)^2 \) where \(\alpha\) and \(\beta\) are the roots of the given equation: \( 2x^2 + 2(m+n)x + (m^2 + n^2) = 0 \)

Step 2: Find the sum and product of the given roots

The sum of the roots of the given equation is: \(\alpha + \beta = \frac{-2(m+n)}{2} = -(m+n)\) The product of the roots of the given equation is: \(\alpha \beta = \frac{m^2 + n^2}{2}\) Now, we have the sum and product of the roots of the given equation.

Step 3: Find the sum and product of the roots of the new equation

The sum of the roots of the new equation is: \(r_1 + r_2 = (\alpha + \beta)^2 + (\alpha - \beta)^2\) The product of the roots of the new equation is: \(r_1 \cdot r_2 = (\alpha + \beta)^2 \cdot (\alpha - \beta)^2\)

Step 4: Express the sum and product of the roots of the new equation in terms of the given roots

Substitute the sum and product of the given roots into the expressions for the sum and product of the roots of the new equation: Sum of roots of the new equation: \(r_1 + r_2 = (\alpha + \beta)^2 + (\alpha - \beta)^2 = [-(m + n)]^2 + [-(m - n)]^2 = m^2 + 2mn + n^2 + m^2 - 2mn + n^2 = 2m^2 + 2n^2\) Product of roots of the new equation: \(r_1 \cdot r_2 = (\alpha + \beta)^2 \cdot (\alpha - \beta)^2 = [-(m + n)]^2 \cdot [-(m - n)]^2 = (m^2 - 2mn + n^2)(m^2 + 2mn + n^2) = (m^2 - n^2)^2\) Now, we have the sum and product of the roots of the new equation in terms of the given roots.

Step 5: Find the equation with the new roots

Using the sum and product of the roots of the new equation, we can find the equation itself: \(x^2 - (r_1 + r_2)x + r_1 \cdot r_2 = x^2 - (2m^2 + 2n^2)x + (m^2 - n^2)^2\) So, the equation with the new roots is: \(x^2 - 4mnx - (m^2 - n^2)^2 = 0\)

Key Concepts

Roots of quadratic equations

Quadratic equations are polynomial equations of degree two, generally expressed in the form \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). The solutions for \( x \) are called the "roots". These can be real or complex numbers, depending on the values of the coefficients.
To find the roots, we often use the quadratic formula:

• \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

The expression under the square root, \( b^2 - 4ac \), is known as the discriminant, which determines the nature of the roots:

• If the discriminant is positive, there are two distinct real roots.
• If it is zero, there is exactly one real root (a repeated root).
• If it is negative, there are two complex roots.

In the given problem, the roots \( \alpha \) and \( \beta \) of the quadratic equation \( 2x^2 + 2(m+n)x + (m^2 + n^2) = 0 \) need to be transformed to find the new roots \( (\alpha + \beta)^2 \) and \( (\alpha - \beta)^2 \). Understanding these transformations helps uncover new relationships between solutions of polynomial equations.

Sum and product of roots

The sum and product of the roots of a quadratic equation are important properties that derive from its coefficients. For a standard quadratic equation \( ax^2 + bx + c = 0 \), the relationships are as follows:

• Sum of roots: \( \alpha + \beta = -\frac{b}{a} \)
• Product of roots: \( \alpha \beta = \frac{c}{a} \)

In the specific problem, \( \alpha + \beta = -(m+n) \) and \( \alpha \beta = \frac{m^2 + n^2}{2} \) are the calculated sum and product for the original quadratic equation. These foundational relations allow us to easily find new equations or transform existing equations by leveraging the known roots.
The task involved in the problem is to find a new quadratic with roots \( (\alpha + \beta)^2 \) and \( (\alpha - \beta)^2 \). For this, the sum is found as:

• \( r_1 + r_2 = (\alpha + \beta)^2 + (\alpha - \beta)^2 = 2m^2 + 2n^2 \)

The product is:

• \( r_1 \cdot r_2 = (m^2 - n^2)^2 \)

These reveals how understanding the sum and product of the initial roots enables deriving equations with desired roots swiftly and accurately.

Polynomial equations

Polynomial equations, such as quadratics, involve expressions that feature powers of the variable. The degree of the polynomial is the highest power of the variable present. A quadratic equation, with a degree of 2, is the simplest form of a polynomial equation where we often need to find its roots by solving for the value of the variable.
Understanding how to manipulate these polynomial equations lets us explore more complex mathematical concepts or derive new equations from existing solutions. For instance, the transformation of roots from \( \alpha \) and \( \beta \) to \( (\alpha+\beta)^2 \) and \( (\alpha-\beta)^2 \) shows how deeper algebraic insights can lead us to comprehensive solutions.
Using these insights, the new polynomial is derived as:

• \( x^2 - 4mnx - (m^2 - n^2)^2 = 0 \)

This illustrates the elegance and power of polynomial equations when solving complex algebraic problems and shows how they are pivotal in various mathematical applications from simple problems to more advanced topics like calculus.