Only two, not three

Let ${ x }^{ 3 }\quad +\quad 5{ x }^{ 2 }\quad +\quad px\quad +\quad q\quad =\quad 0\quad ->\quad eq(1)\quad$

${ x }^{ 3 }\quad +\quad { x }^{ 2 }\quad +\quad px\quad +\quad r\quad =\quad 0\quad ->\quad eq(2)\quad$

Let the 2 common roots be $\alpha \quad ,\quad \beta$ .

Therefore, the roots of eq(1) are $\alpha \quad ,\quad \beta \quad ,\quad { x }_{ 1 }$

Similarly, the roots of eq(2) are $\alpha \quad ,\quad \beta \quad ,\quad { x }_{ 2 }$

Subtracting eq(2) from eq(1), we get

$4{ x }^{ 2 }+(q-r)=0$ -> eq(3)

This equation has the roots $\alpha \quad ,\quad \beta$ .

[Note: We are able to subtract eq(1) and eq(2), only because we have substituted $\alpha \quad ,\quad \beta$ in the given equations in place of 'x' . Hence the roots of eq(3) are $\alpha \quad ,\quad \beta$ ]

From eq(3), we get

$\alpha +\beta$ = 0

$\alpha +\beta + \quad { x }_{ 1 }$ = -5

$\quad { x }_{ 1 } = -5$

$\alpha +\beta + \quad { x }_{ 2 }$ = -1

$\therefore \quad { x }_{ 2 } = -1$

$\therefore \quad { x }_{ 1 }+{ x }_{ 2 }\quad =\quad -6$