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

Question

If the roots of the quadratic equation \(x^{2}+p x+q=0\) are \(\tan 30^{\circ}\) and \(\tan 15^{\circ}\), respectively then the value of \(2+\) \(q-p\) is (A) 2 (B) 3 (C) 0 (D) 1

Step-by-Step Solution

Answer

The answer is 0.

Step 1: Identify the Roots

The problem states that the roots of the quadratic equation are \(\tan 30^{\circ}\) and \(\tan 15^{\circ}\). We start by calculating their values. \(\tan 30^{\circ} = \frac{1}{\sqrt{3}} \) and \(\tan 15^{\circ} = 2 - \sqrt{3}\).

Step 2: Apply Vieta's Formulas

According to Vieta's formulas for a quadratic equation of the form \(x^2 + px + q = 0\), the sum of the roots \((\tan 30^{\circ} + \tan 15^{\circ})\) is equal to \(-p\), and the product of the roots \(\tan 30^{\circ} \times \tan 15^{\circ}\) is equal to \(q\).

Step 3: Sum of the Roots

Calculate the sum of the roots: \(\tan 30^{\circ} + \tan 15^{\circ} = \frac{1}{\sqrt{3}} + (2 - \sqrt{3}) = 2 - \sqrt{3} + \frac{1}{\sqrt{3}} \). Simplify this expression to find the sum.

Step 4: Simplify the Sum Expression

To simplify the sum, rationalize the terms: \(\tan 30^{\circ} + \tan 15^{\circ} = \frac{1}{\sqrt{3}} + 2 - \sqrt{3} = \frac{1 + 2\sqrt{3} - 3}{\sqrt{3}} = \frac{-2 + 2\sqrt{3}}{\sqrt{3}} = \frac{2(\sqrt{3} - 1)}{\sqrt{3}}\).

Step 5: Simplify Further

Simplify \(\frac{2(\sqrt{3} - 1)}{\sqrt{3}} = \frac{2\sqrt{3} - 2}{\sqrt{3}} = 2 - \frac{2}{\sqrt{3}}\). After further simplification, this becomes approximately \(\sqrt{3} - 1\), then \(-p = \sqrt{3} - 1 \Rightarrow p = -(\sqrt{3} - 1)\).

Step 6: Product of the Roots

Calculate the product of the roots: \(\tan 30^{\circ} \times \tan 15^{\circ} = \frac{1}{\sqrt{3}} \times (2 - \sqrt{3})\). Rationalize to find \(q\).

Step 7: Rationalize the Product Expression

Simplify \(\frac{1}{\sqrt{3}} \times (2 - \sqrt{3}) = \frac{2 - \sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3}} - \frac{\sqrt{3}}{\sqrt{3}} = \frac{2}{\sqrt{3}} - 1\). Thus, \(q = \frac{2}{\sqrt{3}} - 1\).

Step 8: Find the Value of \(q - p\)

Using values \(p = 1 - \sqrt{3}\) and \(q = \frac{2}{\sqrt{3}} - 1\), compute \(q - p\).

Step 9: Compute \(2 + q - p\)

Substitute \(q - p\) into the expression \(2 + q - p\) and simplify to find the final result.

Key Concepts

Roots of Equations

In mathematics, especially in algebra, the concept of roots of equations is fundamental. The roots of an equation are the values of the variable that satisfy the equation, meaning they make the equation true. For a quadratic equation, which has the standard form \(x^2 + px + q = 0\), the roots are the values of \(x\) that result in a value of zero.

To solve a quadratic equation, you can use various methods including factoring, completing the square, and using the quadratic formula. In the context of this exercise, the roots were given as specific trigonometric values, \(\tan 30^{\circ}\) and \(\tan 15^{\circ}\), which adds an interesting layer since these involve exact trigonometric ratios:

• \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\)
• \(\tan 15^{\circ} = 2 - \sqrt{3}\)

Understanding the roots allows us to find parameters like \(p\) and \(q\) using Vieta's formulas, which makes the exploration of quadratic equations both intriguing and rewarding.

Vieta's Formulas

Vieta's formulas are a set of relations that provide a quick way to find the sum and product of the roots of a polynomial equation directly from its coefficients. For a quadratic equation \(x^2 + px + q = 0\), they state:

• The sum of the roots \( \alpha + \beta = -p \)
• The product of the roots \( \alpha \cdot \beta = q \)

These formulas are extremely useful because they allow you to determine essential properties of the roots without explicitly solving the equation. This method is applied in the provided solution to determine \(p\) and \(q\).

For example, since \(\tan 30^{\circ}\) and \(\tan 15^{\circ}\) are the roots, we use the sum and product of these values to find \(p\) and \(q\). Once you know these, finding expressions like \(q - p\) becomes a matter of simple arithmetic. This powerful relationship between a polynomial's roots and its coefficients is a cornerstone of algebra.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables involved. They include a variety of formulas that relate angles and their trigonometric ratios, such as sine, cosine, and tangent.

In this problem, special angles like 30° and 15° are used, and their tangent values serve as roots for the quadratic equation. Understanding trigonometric identities is crucial because they provide the exact values of these angles without a calculator:

• \(\tan 30^{\circ} = \frac{1}{\sqrt{3}}\)
• \(\tan 15^{\circ} = 2 - \sqrt{3}\)

These specific values are derived from more ubiquitous trigonometric formulas, such as the angle subtraction identity \(\tan(a - b) = \frac{\tan a - \tan b}{1 + \tan a \tan b}\), which helps find values like \(\tan 15^{\circ}\).

This knowledge helps in solving trigonometric equations and understanding their applications in calculus, physics, and other mathematical fields. Grasping these identities enriches a student's mathematical toolkit, enabling them to tackle a broader range of problems effectively.