Question:

**Step 1: Simplify the first equation** The first given equation is $(6x^2 + 3) - rx + 2x^2 - 1 = 0$. Combining like terms, we obtain: $$ 8x^2 - rx + 2 = 0 \quad \text{(Equation 1)} $$ **Step 2: Simplify the second equation** The second given equation is $6(2x^2 + 1) + px + 4x^2 - 2 = 0$. Distributing and combining like terms, we obtain: $$ 12x^2 + 6 + px + 4x^2 - 2 = 0 $$ $$ 16x^2 + px + 4 = 0 \quad \text{(Equation 2)} $$ **Step 3: Apply the condition for common roots** If two quadratic equations $A_1x^2 + B_1x + C_1 = 0$ and $A_2x^2 + B_2x + C_2 = 0$ have both roots common, then their coefficients must be proportional. That is: $$ \frac{A_1}{A_2} = \frac{B_1}{B_2} = \frac{C_1}{C_2} $$ Applying this condition to Equation 1 and Equation 2: $$ \frac{8}{16} = \frac{-r}{p} = \frac{2}{4} $$ **Step 4: Determine the relationship between $r$ and $p$** From the proportionality established in Step 3, we can equate the ratios: $$ \frac{1}{2} = \frac{-r}{p} $$ Cross-multiplying yields the relationship between $p$ and $r$: $$ p = -2r $$ **Step 5: Evaluate the expression $2r - p$** Substitute the relationship $p = -2r$ into the expression $2r - p$: $$ 2r - p = 2r - (-2r) $$ $$ 2r - p = 2r + 2r $$ $$ 2r - p = 4r $$ For the expression $2r-p$ to be a unique numerical value, $r$ must be uniquely determined. This occurs when $4r=0$, which implies $r=0$. If $r=0$, then from $p = -2r$, we have $p = -2(0) = 0$. Substituting $r=0$ and $p=0$ into the expression $2r-p$: $$ 2(0) - 0 = 0 $$ Thus, $2r - p = 0$.