The equations $3x^2-5x+p=0$ and $2x^2-2x+q=0$ have one common root. The sum of the other roots of this equa

The correct option is:
$\frac{8}{3}-p+\frac{3}{2}q$

Step-by-Step Explanation:

Step 1: Define the roots of both equations.
Let the common root of both equations be $\alpha$.
Let the second root of the first equation $3x^2-5x+p=0$ be ${\beta }_1$.
Let the second root of the second equation $2x^2-2x+q=0$ be ${\beta }_2$.

Step 2: Use the relationship between roots and coefficients.
For the first quadratic equation, the sum of roots is given by:
$\alpha +{\beta }_1=\frac{5}{3}\Rightarrow {\beta }_1=\frac{5}{3}-\alpha$

For the second quadratic equation, the sum of roots is given by:
$\alpha +{\beta }_2=\frac{2}{2}=1\Rightarrow {\beta }_2=1-\alpha$

Step 3: Express the sum of the non-common roots.
We need to find the sum of the other roots, ${\beta }_1+{\beta }_2$:
${\beta }_1+{\beta }_2=(\frac{5}{3}-\alpha )+(1-\alpha )$
${\beta }_1+{\beta }_2=\frac{8}{3}-2\alpha$ --- (Equation 1)

Step 4: Find the value of the common root $\alpha$.
Since $\alpha$ is a common root, it satisfies both given equations:
1) $3{\alpha }^2-5\alpha +p=0$
2) $2{\alpha }^2-2\alpha +q=0$

To eliminate the ${\alpha }^2$ term, multiply the first equation by 2 and the second equation by 3:
$6{\alpha }^2-10\alpha +2p=0$
$6{\alpha }^2-6\alpha +3q=0$

Subtracting the second modified equation from the first:
$(6{\alpha }^2-10\alpha +2p)-(6{\alpha }^2-6\alpha +3q)=0$
$-4\alpha +2p-3q=0$
$4\alpha =2p-3q\Rightarrow \alpha =\frac{2p-3q}{4}$

Step 5: Substitute $\alpha$ into Equation 1.
${\beta }_1+{\beta }_2=\frac{8}{3}-2\left(\frac{2p-3q}{4}\right)$
${\beta }_1+{\beta }_2=\frac{8}{3}-\frac{2p-3q}{2}$
${\beta }_1+{\beta }_2=\frac{8}{3}-p+\frac{3}{2}q$

Thus, the sum of the other roots of the given equations is $\frac{8}{3}-p+\frac{3}{2}q$.