Key Concepts and Formulas

• Vieta's Formulas: For a quadratic equation $ax^2 + bx + c = 0$ with roots $\alpha$ and $\beta$, we have $\alpha + \beta = -\frac{b}{a}$ and $\alpha\beta = \frac{c}{a}$.
• Ratio of Roots: If $\lambda = \frac{\alpha}{\beta}$, then $\frac{1}{\lambda} = \frac{\beta}{\alpha}$.
• Algebraic Identity: $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$.

Step-by-Step Solution

Step 1: Understand the condition on the ratio of roots and relate it to the sum and product of roots.

We are given that $\lambda + \frac{1}{\lambda} = 1$, where $\lambda = \frac{\alpha}{\beta}$. Substituting this into the given equation, we get: $\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = 1$ Multiplying both sides by $\alpha\beta$, we obtain: $\alpha^2 + \beta^2 = \alpha\beta$ Now, we use the algebraic identity $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$ to relate this to the sum and product of roots. Substituting $\alpha^2 + \beta^2 = \alpha\beta$ into the identity, we get: $(\alpha + \beta)^2 = \alpha\beta + 2\alpha\beta$ $(\alpha + \beta)^2 = 3\alpha\beta$ This equation connects the sum and product of the roots, which we can find using Vieta's formulas.

Step 2: Apply Vieta's formulas to the given quadratic equation.

The given quadratic equation is $3m^2 x^2 + m(m – 4)x + 2 = 0$. For this equation to be a quadratic equation in $x$, we must have $3m^2 \neq 0$, which means $m \neq 0$. Here, $a = 3m^2$, $b = m(m-4)$, and $c = 2$. Using Vieta's formulas:

• Sum of roots: $\alpha + \beta = -\frac{b}{a} = -\frac{m(m-4)}{3m^2}$. Since $m \neq 0$, we can simplify this to $\alpha + \beta = -\frac{m-4}{3m}$.
• Product of roots: $\alpha\beta = \frac{c}{a} = \frac{2}{3m^2}$.

Step 3: Substitute Vieta's formulas into the derived relationship.

Now, we substitute the expressions for $(\alpha + \beta)$ and $\alpha\beta$ into the relationship $(\alpha + \beta)^2 = 3\alpha\beta$: $\left(-\frac{m-4}{3m}\right)^2 = 3 \left(\frac{2}{3m^2}\right)$ Simplify both sides: $\frac{(m-4)^2}{9m^2} = \frac{6}{3m^2} = \frac{2}{m^2}$

Step 4: Solve for m.

Since $m \neq 0$, we can multiply both sides of the equation by $9m^2$: $(m-4)^2 = 18$ Take the square root of both sides: $m-4 = \pm\sqrt{18} = \pm 3\sqrt{2}$ Solving for $m$: $m = 4 \pm 3\sqrt{2}$ This gives two possible values for $m$: $4 + 3\sqrt{2}$ and $4 - 3\sqrt{2}$.

Step 5: Determine the least value of m.

We need to find the least value of $m$. Comparing the two values: $4 + 3\sqrt{2} > 0$ $4 - 3\sqrt{2} = 4 - 3(1.414...) \approx 4 - 4.242 = -0.242 < 0$ Therefore, the least value of $m$ is $4 - 3\sqrt{2}$.

Common Mistakes & Tips

• Checking for $m \neq 0$: Always remember to check that $m \neq 0$ since it's in the denominator of the expressions for the sum and product of the roots.
• Sign Errors: Be careful with signs when substituting and simplifying expressions, especially when squaring terms.

Summary

We started by relating the given condition on the ratio of roots to the sum and product of roots. Applying Vieta's formulas to the given quadratic equation allowed us to express the sum and product of roots in terms of $m$. Substituting these expressions into the derived relationship and solving for $m$, we found two possible values: $4 + 3\sqrt{2}$ and $4 - 3\sqrt{2}$. The least of these is $4 - 3\sqrt{2}$.

Final Answer

The final answer is \boxed{4 - 3\sqrt{2}}, which corresponds to option (B).