If λ be the ratio of the roots of the quadratic equation in x, 3m²x²+m(m-4)x+2 = 0, then the least v

Q: Why can't we just solve for $\lambda$ from $\lambda + 1/\lambda = 1$?

If you solve $\lambda^2 - \lambda + 1 = 0$, you get complex roots. While mathematically valid, it complicates the problem unnecessarily. By keeping the ratio as a variable, we leverage the symmetry of the roots, which is a much more elegant and efficient path.

Q: Why do we assume $m \neq 0$?

Look at the original equation $3m^2x^2 + m(m-4)x + 2 = 0$. If $m=0$, the $x^2$ and $x$ terms vanish, leaving $2=0$, which is impossible. Thus, $m$ cannot be zero, allowing us to safely divide by $m^2$.

Q: How do I know which value of $m$ to choose?

The question asks for the 'least value'. After solving $(m-4)^2 = 18$, we get $m = 4 \pm 3\sqrt{2}$. Since $3\sqrt{2} > 0$, $4 - 3\sqrt{2}$ is clearly smaller than $4 + 3\sqrt{2}$.

Question

If $\lambda$ be the ratio of the roots of the quadratic equation in $x$, $3m^2x^2+m(m-4)x+2 = 0$, then the least value of $m$ for which $\lambda + 1/\lambda = 1$, is :

Options:

(A) $2-\sqrt{3}$

(B) $4-3\sqrt{2}$

(C) $-2+\sqrt{2}$

(D) $4-2\sqrt{3}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before this quadratic equation: $3m^2x^2 + m(m-4)x + 2 = 0$. It looks like a standard, perhaps even intimidating, algebraic expression.

In the world of JEE Advanced, we don't just solve equations; we decode them. We are given a ratio of roots $\lambda = \frac{\alpha}{\beta}$ and a condition $\lambda + \frac{1}{\lambda} = 1$. This is not just a constraint; it is a hidden symmetry.

### Decoding the Ratio Condition

Let's dive into the algebra. If $\lambda = \frac{\alpha}{\beta}$, then the condition $\lambda + \frac{1}{\lambda} = 1$ becomes $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$.

When you take the LCM, you get $\frac{\alpha^2 + \beta^2}{\alpha\beta} = 1$, which simplifies beautifully to:

$$\alpha^2 + \beta^2 = \alpha\beta$$

This is the bridge between the ratio condition and the coefficients of our quadratic equation.

### The Algebraic Bridge

Now, we need to connect this to the sum and product of the roots. We know the identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$.

Substituting our relation $\alpha^2 + \beta^2 = \alpha\beta$ into this identity, we get $(\alpha + \beta)^2 = \alpha\beta + 2\alpha\beta$, which simplifies to:

$$(\alpha + \beta)^2 = 3\alpha\beta$$

This is a powerful, elegant result. It allows us to bypass the individual roots and work directly with the coefficients.

### The Final Calculation

From our original equation $3m^2x^2 + m(m-4)x + 2 = 0$, we use Vieta's formulas. The sum of roots is $\alpha + \beta = -\frac{m(m-4)}{3m^2} = -\frac{m-4}{3m}$, and the product of roots is $\alpha\beta = \frac{2}{3m^2}$.

Substituting these into our relation $(\alpha + \beta)^2 = 3\alpha\beta$, we get:

$$\left( -\frac{m-4}{3m} ight)^2 = 3 \left( \frac{2}{3m^2} ight)$$

Expanding the left side gives:

$$\frac{(m-4)^2}{9m^2} = \frac{2}{m^2}$$

Since $m 
eq 0$, we can cancel $m^2$ to get $(m-4)^2 = 18$. Taking the square root, $m-4 = \pm 3\sqrt{2}$, so $m = 4 \pm 3\sqrt{2}$.

The **least value** is **$4 - 3\sqrt{2}$**. This journey shows that even complex-looking problems have a simple, elegant core if you know where to look.

Animated Solution for Mathematics - Quadratic Equations: If $λ$ be the ratio of the roots of the quadratic equation in $x$ , $3 m^{2} x^{2} + m (m - 4) x + 2 = 0$ , then the least value of $m$ for which $λ + 1/ λ = 1$ , is :

Select Answer:

Visualized Solution

The Quadratic Equation

Condition on Roots

Substituting $λ$

Simplifying the Equation

Using Algebraic Identities

Sum and Product of Roots

Forming the Equation in $m$

Simplifying the $m$ Equation

Solving for $m$

Roots for $m$

Finding the Least Value

The Sigma Insight: Relation Between Roots and Coefficients

Analyzing the Setup

Decoding the Ratio Condition

The Algebraic Bridge

The Final Calculation

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Select Answer:

Visualized Solution

The Quadratic Equation

Condition on Roots

Substituting .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } λ

Simplifying the Equation

Using Algebraic Identities

Sum and Product of Roots

Forming the Equation in .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Simplifying the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m

Roots for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } m