Question

The number of real roots of the equation \(x^{3}-\frac{1}{x^{3}}+4\left(x-\frac{1}{x}\right)=0, x \neq 0\) (a) 3 (b) 0 (c) 1 (d) 2

Short answer

(c) 1 real root

Step-by-step solution

Rewrite the given equation to isolate 'x'

Multiply both sides of the equation by \(x^3\) to get rid of the fraction: \(x^3(x^3 - \frac{1}{x^3}) + 4x^3(x - \frac{1}{x}) = 0\) Now, we have: \(x^6 - 1 + 4x^4(x - \frac{1}{x}) = 0\)

Simplify the equation further using the difference of squares

Observe that \((x^6 - 1)\) is a difference of two squares and can be written as: \((x^3-1)(x^3+1)+4x^4(x-\frac{1}{x})=0\)

Factor out \(x-1\) and rewrite the equation

To simplify the expression inside the parenthesis, we need to factor out \((x-1)\) and rewrite the equation: \((x-1)(x^2+x+1)(x^3+1)+4x^4(x-\frac{1}{x})=0\)

Factor out \(x\) and rewrite the equation

Now, factor out \(x\) from the second term and rewrite the equation: \((x-1)(x^2+x+1)(x^3+1)+4x^5(1-\frac{1}{x^2})=0\)

Multiply both sides by \(x^2\) and rewrite the equation

Multiply both sides by \(x^2\) to eliminate the remaining fraction: \((x-1)(x^2+x+1)(x^3+1)x^2 + 4x^5(1-\frac{1}{x^2})x^2 = 0\) This simplifies to: \((x-1)(x^2+x+1)(x^5+x^3)+4x^7(1-x^{-2})=0\) After combining terms, we get: \((x-1)(x^7+x^5+x^3)+4x^7-4=0\)

Find the possible number of real roots

Observe that the polynomial inside the parenthesis is increasing, and its coefficient of \(x^7\) is positive. Also, note that the polynomial \((x-1)\) has one root at \(x=1\). Therefore, the equation \((x-1)(x^7+x^5+x^3)+4x^7-4=0\) will have at least one real root. From the given options, we can see that the answer is: (c) 1 real root.

Key concepts

Polynomial Factorization

Understanding how to factorize polynomials is a fundamental concept in algebra and is crucial for solving equations efficiently. Factorization involves expressing a polynomial as a product of its factors, much like breaking down a number into its prime components. The aim is to rewrite the polynomial in a way that reveals its roots or simplifies further manipulation.

In the context of the given exercise, factorization was used to simplify the polynomial equation by recognizing the difference of squares and factoring out common terms like \(x-1\). If a polynomial is completely factorized, it becomes much easier to find its solutions or roots, which are the values of \(x\) that satisfy the equation. For IIT-JEE aspirants, mastering polynomial factorization is a must as it regularly appears in problems across the syllabus.

Difference of Squares

A difference of squares is a specialized algebraic form that students encounter frequently. It is an expression of the form \(a^2 - b^2\), which can be factored into \(a + b)(a - b)\). Recognizing this pattern allows for quick factorization of more complicated expressions.

In our exercise, we identified \(x^6 - 1\) as a difference of squares. By applying the formula, we were able to break the term down into \(x^3 + 1\) and \(x^3 - 1\), further simplifying the problem. Spotting a difference of squares is an invaluable skill for those preparing for mathematics-intensive exams like the IIT-JEE, as it frequently enables simplification of what might initially seem to be daunting expressions.

Cubic Equations

Cubic equations are polynomial equations that involve an \(x^3\) term and have the general form \(ax^3 + bx^2 + cx + d = 0\). While they are more complex than quadratic equations, they can still be solved by a variety of methods, including factorization, synthetic division, and using the cubic formula.

The given exercise included transforming part of the equation into a cubic factor, which was then more easily handled through further factorization. Becoming comfortable with manipulating cubic equations is important for higher-level math and for tackling competitive exams like the IIT-JEE, where understanding the behavior of polynomials of all degrees can give a strategic edge.

IIT-JEE Mathematics Preparation

Preparing for the IIT-JEE Mathematics paper requires strategic planning and a thorough understanding of concepts, such as polynomial equations, including cubic equations and factorization techniques. Successful candidates often practice identifying and applying algebraic identities, like the difference of squares, in complex problems.

To excel in IIT-JEE Mathematics, students should focus on conceptual clarity, problem-solving speed, and accuracy. They should regularly solve a variety of problems, including those in their textbooks and from additional resources, to become adept at recognizing patterns, such as the ones seen in the provided exercise. Getting comfortable with a wide range of problems will help students tackle the IIT-JEE's challenging and creative questions with confidence.