If bot the roots of `lamda(6x^(2)+3)+rx+2x^(2)-1=0` and `6 lamda(2x^(2)+1)+px+4x^(2)-2=0` are common, then `2r-p` is equal to

Options

A. -1
B. 0
C. 1
D. 2

Answer

To solve the problem, we need to analyze the two given equations and find the relationship between \( r \) and \( p \) based on the condition that they have common roots.

### Step 1: Write the equations in standard form
The first equation is given as:
\[ \lambda(6x^2 + 3) + rx + 2x^2 - 1 = 0 \]
This can be rearranged as:
\[ (6\lambda + 2)x^2 + rx + (3\lambda - 1) = 0 \]

The second equation is:
\[ 6\lambda(2x^2 + 1) + px + 4x^2 - 2 = 0 \]
This can be rearranged as:
\[ (12\lambda + 4)x^2 + px + (6\lambda - 2) = 0 \]

### Step 2: Set up the condition for common roots
For the two quadratic equations to have common roots, the ratios of their coefficients must be equal:
\[ \frac{6\lambda + 2}{12\lambda + 4} = \frac{r}{p} = \frac{3\lambda - 1}{6\lambda - 2} \]

### Step 3: Simplify the first ratio
The first ratio can be simplified:
\[ \frac{6\lambda + 2}{12\lambda + 4} = \frac{2(3\lambda + 1)}{4(3\lambda + 1)} = \frac{1}{2} \quad \text{(for } 3\lambda + 1 \neq 0\text{)} \]

### Step 4: Set up the equation from the first ratio
From the first ratio, we have:
\[ \frac{r}{p} = \frac{1}{2} \implies 2r = p \]

### Step 5: Substitute into the second ratio
Now, substituting \( p = 2r \) into the second ratio:
\[ \frac{r}{2r} = \frac{1}{2} \]
This is consistent and confirms our earlier finding.

### Step 6: Find \( 2r - p \)
Now, we can find \( 2r - p \):
\[ 2r - p = 2r - 2r = 0 \]

### Conclusion
Thus, the value of \( 2r - p \) is:
\[ \boxed{0} \]