If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root then which of the following is true If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root then which of the following is true:

$1$. $p=q$

$2$. $p+q+1=0$

$3$. $p+q=0$

$4$. $\textrm {both} 1,2$.

MY WORK:

$$x=\dfrac {-p\pm \sqrt {p^2-4q}}{2}$$ 
And,
$$x=\dfrac {-q\pm \sqrt {q^2-4p}}{2}$$

HOW DO I GET IT COMPLETED?
If $$x^2+px+q = 0 = x^2+qx+p$$ then $$px+q=qx+p$$ So either $p=q$ or $x=1$ and thus $1+p+q=0$.

So one of (1) or (2) must be true (it is not necessary for both to be true.) It's not clear what "both" is doing in the question
$$a_1x^2+b_1x+c_1=0$$
$$a_2x^2+b_2x+c_2=0$$.

Condition for one root to be common of two quadratic equations is:
$$(c_1a_2-c_2a_1)^2=(b_1c_2-b_2c_1)(a_1b_2-a_2b_1)$$

In your case it is:
$$(q-p)^2=(p^2-q^2)(q-p)$$
$$(q-p)=(p+q)(p-1)$$
$$p+q=-1$$
$$p+q+1=0$$

So the correct option would be $2$.