If the roots of the quadratic equation x² + px + q = 0 are tan 30° and tan 15°, respectively, then t

Q: Wait, why do we use Vieta's formulas instead of just calculating $\tan 30^\circ$ and $\tan 15^\circ$ directly?

That is a great question! While you could calculate $\tan 30^\circ = \frac{1}{\sqrt{3}}$ and $\tan 15^\circ = 2 - \sqrt{3}$, it would lead to messy radical expressions. Using Vieta's formulas and trigonometric identities is much more elegant, less prone to calculation errors, and teaches you to see the underlying structure of the problem.

Q: How did you know to use the $\tan(A + B)$ formula?

Look at the structure of the expression we need to find: $2 + q - p$. We know $q$ is the product of the roots and $-p$ is the sum of the roots. The tangent addition formula $\tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B}$ is the perfect bridge because it contains both the sum (in the numerator) and the product (in the denominator) of the tangents. It is a classic JEE pattern to link algebra and trigonometry this way.

Q: Is there a sign error in the Vieta's formula? I thought the sum of roots was $p$.

Remember, for a quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$. In our equation $x^2 + px + q = 0$, the coefficient of $x$ is $p$, so the sum of the roots is $-\frac{p}{1} = -p$. Always be careful with that negative sign; it is one of the most common traps in JEE exams!

Question

If the roots of the quadratic equation $x^2 + px + q = 0$ are $	an 30^\circ$ and $	an 15^\circ$, respectively, then the value of $2 + q - p$ is

Options:

(A) 2

(B) 3

(C) 0

(D) 1

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, fellow traveler on the path to JEE mastery. Today, we are not just solving a quadratic equation; we are witnessing a beautiful dance between two distinct worlds: the world of algebraic coefficients and the world of trigonometric angles.

When you look at the equation $x^2 + px + q = 0$, do not just see a collection of symbols. See a structure waiting to be unlocked. We are given the roots as $\alpha = 	an 30^\circ$ and $\beta = 	an 15^\circ$.

Our mission is to find the value of $2 + q - p$. This is not a brute-force calculation; it is a puzzle of relationships.

### The Vieta Connection

Our first step is to invoke the wisdom of Vieta. For any quadratic equation $ax^2 + bx + c = 0$, the sum of the roots is $-\frac{b}{a}$ and the product is $\frac{c}{a}$.

In our specific case, $a = 1$, $b = p$, and $c = q$. Therefore, the sum of our roots is:

$$	an 30^\circ + 	an 15^\circ = -p$$

The product of our roots is:

$$	an 30^\circ \cdot 	an 15^\circ = q$$

These two relations are the keys to our kingdom. We have successfully translated the roots into algebraic expressions involving $p$ and $q$.

### The Trigonometric Bridge

Now, we stand at a crossroads. We have $p$ and $q$, but how do we connect them to the expression $2 + q - p$?

This is where the beauty of trigonometry shines. We have the sum of two tangents and the product of the same two tangents. Recall the compound angle formula for tangent:

$$	an(A + B) = \frac{	an A + 	an B}{1 - 	an A 	an B}$$

This formula is a masterpiece of design. It places the sum of the tangents in the numerator and the product in the denominator. It is as if the formula was written specifically for this problem.

### The Algebraic Dance

Let us set $A = 30^\circ$ and $B = 15^\circ$. The sum of these angles is $30^\circ + 15^\circ = 45^\circ$. We know that $	an 45^\circ = 1$.

Now, substitute our Vieta relations into the formula:

$$1 = \frac{-p}{1 - q}$$

This is the moment of transformation. We have moved from the realm of angles to the realm of pure algebra. Cross-multiplying gives us:

$$1 - q = -p$$

Rearranging this, we find $q - p = 1$. This is the hidden truth we were searching for.

### The Final Victory

We are almost there. The problem asks for the value of $2 + q - p$. We now know that $q - p = 1$.

Substituting this into our target expression, we get:

$$2 + 1 = 3$$

And there it is! The complexity dissolves, leaving behind a simple, elegant integer. The final answer is **3**.

Animated Solution for Mathematics - Trigonometry: If the roots of the quadratic equation $x^{2} + p x + q = 0$ are $tan 3 0^{\circ}$ and $tan 1 5^{\circ}$ , respectively, then the value of $2 + q - p$ is

Select Answer:

Visualized Solution

Identify the Quadratic Equation and its Roots

Sum of Roots using Vieta's Formula

Product of Roots using Vieta's Formula

The Trigonometric Bridge

Evaluate the Compound Angle

Algebraic Substitution

Rearranging the Equation

Find the Final Value

The Sigma Insight: Trigonometric Functions of Compound Angles

Analyzing the Setup

The Vieta Connection

The Trigonometric Bridge

The Algebraic Dance

The Final Victory

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Animated Solution for Mathematics - Trigonometry: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the roots of the quadratic equation x2+px+q=0 are tan30∘ and tan15∘, respectively, then the value of 2+q−p is

Select Answer:

Visualized Solution

Identify the Quadratic Equation and its Roots

Sum of Roots using Vieta's Formula

Product of Roots using Vieta's Formula

The Trigonometric Bridge

Evaluate the Compound Angle

Algebraic Substitution

Rearranging the Equation

Find the Final Value

The Sigma Insight: Trigonometric Functions of Compound Angles

Analyzing the Setup

The Vieta Connection

The Trigonometric Bridge

The Algebraic Dance

The Final Victory

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