If the equation ${x^2} + px + q = 0$ and ${x^2} + qx + p = 0$, have a common root, then $p + q + 1 = $
A. $0$
B. $1$
C. $2$
D. $ - 1$

Hint: In order to solve this type of question, first we will consider the given equations. Then, we will assume the common root of both the equations and rewrite them. Next, we will subtract both the equations and substitute the obtained value in any of the rewritten equations to get the final answer.

Formula used:
If $\alpha $ is the root of ${x^2} + px + q = 0$ then ${\alpha ^2} + p\alpha + q = 0$.

Complete step by step solution:
We are given the equations,
${x^2} + px + q = 0$
${x^2} + qx + p = 0$
Let the common root of both the equations be $\alpha .$
${\alpha ^2} + p\alpha + q = 0$ ………………..equation$\left( 1 \right)$
${\alpha ^2} + q\alpha + p = 0$ ………………..equation$\left( 2 \right)$
Subtract equation $\left( 2 \right)$ from $\left( 1 \right)$,
$\left( {p - q} \right)\alpha + \left( {q - p} \right) = 0$
$\alpha = \dfrac{{\left( {p - q} \right)}}{{\left( {p - q} \right)}}$
$\alpha = 1$
Substituting this value in equation $\left( 1 \right)$,
${\left( 1 \right)^2} + p\left( 1 \right) + q = 0$
$p + q + 1 = 0$
Thus, the correct option is A.

Note: The sum of roots of a quadratic equation is given by $\dfrac{{ - b}}{a}.$ The product of the roots of a quadratic equation is given by $\dfrac{c}{a}.$ If $\alpha $ is the root of ${x^2} + px + q = 0$ then it will satisfy ${\alpha ^2} + p\alpha + q = 0$ which will be followed by other roots of the quadratic equation also.