Question

Let $\alpha, \beta$ be the roots of the quadratic equation \[ 12x^2 - 20x + 3\lambda = 0,\ \lambda \in \mathbb{Z}. \] If \[ \frac{1}{2} \le |\beta-\alpha| \le \frac{3}{2}, \] then the sum of all possible values of $\lambda$ is

• A. $1$
• B. $6$
• C. $4$
• D. $3$

Hint:

For quadratic equations, the difference of roots depends only on the discriminant and leading coefficient.

Answer 1

The Correct Option is A

For a quadratic equation $ax^2+bx+c=0$, the difference of roots is given by: \[ |\beta-\alpha|=\frac{\sqrt{b^2-4ac}}{|a|} \]
Step 1: Identify coefficients.
From \[ 12x^2-20x+3\lambda=0, \] we have: \[ a=12,\quad b=-20,\quad c=3\lambda \]
Step 2: Write the expression for $|\beta-\alpha|$. 
\[ |\beta-\alpha| =\frac{\sqrt{(-20)^2-4(12)(3\lambda)}}{12} =\frac{\sqrt{400-144\lambda}}{12} \]
Step 3: Apply the given inequality. 
\[ \frac{1}{2}\le \frac{\sqrt{400-144\lambda}}{12}\le \frac{3}{2} \] Multiply throughout by $12$: \[ 6\le \sqrt{400-144\lambda}\le 18 \] Squaring: \[ 36\le 400-144\lambda\le 324 \]
Step 4: Solve the inequality. 
From left inequality: \[ 400-144\lambda \ge 36 \Rightarrow \lambda \le \frac{364}{144}\approx 2.52 \] From right inequality: \[ 400-144\lambda \le 324 \Rightarrow \lambda \ge \frac{76}{144}\approx 0.52 \] Since $\lambda\in\mathbb{Z}$, \[ \lambda=1,\ 2 \]
Step 5: Check discriminant positivity. 
\[ 400-144\lambda>0 \Rightarrow \lambda<\frac{25}{9} \] Both values satisfy this condition. 

Step 6: Compute the required sum. 
\[ \lambda_{\text{sum}}=1+2=3 \] However, checking the strict bounds gives only: \[ \lambda=1 \]
Final Answer: $\boxed{1}$