The harmonic mean of the roots of the equation (5 + √2) x² - (4 + √5) x + 8 + 2 √5 = 0 is (A) 2 (B)

Q: Wait, do I really have to calculate the roots using the quadratic formula? That looks impossible with these square roots!

Absolutely not! That is the classic trap. In JEE problems, if you see a complex expression, there is almost always a symmetry or a property (like Vieta's) that allows you to bypass the brute-force calculation. Always look for the 'shortcut' through the properties of roots.

Q: Why does the Harmonic Mean formula have a 2 in the numerator? Is it always $2\alpha\beta / (\alpha + \beta)$?

Yes, for exactly two numbers, the Harmonic Mean is defined as the reciprocal of the arithmetic mean of the reciprocals. Mathematically, $1/H = (1/\alpha + 1/\beta) / 2$. When you simplify that, you get $2\alpha\beta / (\alpha + \beta)$. It is a beautiful piece of symmetry!

Q: I got stuck because the denominator $(5 + \sqrt{2})$ didn't seem to simplify. Did I make a mistake?

It is very common to feel that way! When you see a common term like $(5 + \sqrt{2})$ appearing in both the numerator and denominator of a complex fraction, trust your algebra. It is a 'gift' from the problem setter meant to simplify the expression. Keep going, and you will see it vanish.

Q: Is there any case where the Harmonic Mean is undefined?

Great observation! The Harmonic Mean is undefined if the sum of the roots $(\alpha + \beta)$ is zero. In this specific quadratic equation, the sum is clearly non-zero, so we are safe. Always keep an eye on the denominator in your final steps!

Q: How do I know when to factor out terms like I did with $8 + 2\sqrt{5}$?

Whenever you have a complex expression, look for common factors that might match other parts of your equation. By factoring out a 2, you revealed $(4 + \sqrt{5})$, which perfectly matched the denominator. It is like finding a hidden key to a locked door.

Question

The harmonic mean of the roots of the equation $(5 + \sqrt{2}) x^2 - (4 + \sqrt{5}) x + 8 + 2 \sqrt{5} = 0$ is

Options:

(A) 2

(B) 4

(C) 6

(D) 8

Sigma Percentile · Accepted Answer

### The Elegance of Hidden Symmetry

Welcome, fellow traveler on the path to JEE mastery. Today, we encounter a problem that, at first glance, looks like a chaotic mess of irrational numbers.

You see $(5 + \sqrt{2})x^2 - (4 + \sqrt{5})x + 8 + 2\sqrt{5} = 0$ and your instinct might be to panic. But here is the secret: in the world of competitive mathematics, the most intimidating expressions often hide the most elegant simplifications.

We are not here to fight the numbers; we are here to dance with them.

### The Harmonic Mean: A Geometric Perspective

We are asked to find the Harmonic Mean ($H.M.$) of the roots $\alpha$ and $\beta$. Recall that the Harmonic Mean of two numbers is defined as:

$$H.M. = \frac{2\alpha\beta}{\alpha + \beta}$$

Notice something profound here? We do not need the individual values of $\alpha$ and $\beta$. We only need their sum and their product.

This is the power of Vieta's formulas. By shifting our focus from the roots themselves to the relationship between the roots and the coefficients, we have already won half the battle.

### Deploying Vieta's Arsenal

Let us compare our given equation to the standard quadratic form $ax^2 + bx + c = 0$. We identify our coefficients as:

$$a = 5 + \sqrt{2}$$

$$b = -(4 + \sqrt{5})$$

$$c = 8 + 2\sqrt{5}$$

Using Vieta's formulas, we know that the sum of the roots is $\alpha + \beta = -\frac{b}{a}$ and the product is $\alpha\beta = \frac{c}{a}$. Substituting our values, we get:

$$\alpha + \beta = \frac{4 + \sqrt{5}}{5 + \sqrt{2}}$$

$$\alpha\beta = \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}}$$

### The Moment of Simplification

Now, let us bring these into our $H.M.$ formula. Watch closely as the complexity begins to dissolve:

$$H.M. = \frac{2 \left( \frac{8 + 2\sqrt{5}}{5 + \sqrt{2}} \right)}{\frac{4 + \sqrt{5}}{5 + \sqrt{2}}}$$

Do you see it? The term $(5 + \sqrt{2})$ is present in both the numerator and the denominator. It is a common factor that acts as a bridge, allowing us to cancel it out entirely.

We are left with:

$$H.M. = \frac{2(8 + 2\sqrt{5})}{4 + \sqrt{5}}$$

### The Final Reveal

We are almost there. Look at the numerator $8 + 2\sqrt{5}$. If we factor out a $2$, we get $2(4 + \sqrt{5})$.

Suddenly, the expression becomes:

$$H.M. = \frac{2 \cdot 2(4 + \sqrt{5})}{4 + \sqrt{5}}$$

The term $(4 + \sqrt{5})$ cancels out perfectly, leaving us with $2 \cdot 2 = 4$.

The final answer is **4**.

### Reflecting on the Journey

Take a moment to appreciate what just happened. We started with a terrifying quadratic equation filled with square roots, and through the systematic application of Vieta's formulas and algebraic intuition, we arrived at a clean, simple integer.

This is the beauty of JEE mathematics. It is not about brute force; it is about recognizing the structure, trusting the process, and watching as the complexity collapses into simplicity.

You have the tools, you have the logic—now go forth and conquer the next challenge with this same confidence.

Animated Solution for Mathematics - Quadratic Equations: The harmonic mean of the roots of the equation $(5 + 2) x^{2} - (4 + 5) x + 8 + 25 = 0$ is

Select Answer:

Visualized Solution

Identify the Equation

Define Harmonic Mean

Extract Coefficients

Sum of Roots $(α + β)$

Product of Roots $(α β)$

Substitute into H.M. Formula

Simplify the Expression

Factorize the Numerator

Final Calculation

The Sigma Insight: Relation Between Roots and Coefficients

The Elegance of Hidden Symmetry