If roots of equation ${ x }^{ 2 }+x+1=0$ are $a$, $b$ and roots of ${ x }^{ 2 }+px+q=0$ are $\dfrac { a }{ b } $, $\dfrac { b }{ a }$; then the value of $p+q$ is

For the equation x^2 + x + 1 = 0, the sum of the roots a + b equals -1 and the product ab equals 1. The roots of x^2 + px + q = 0 are a/b and b/a. The sum of these new roots is a/b + b/a = (a^2 + b^2)/ab, which equals ((a + b)^2 - 2ab)/ab. Substituting the known values gives ((-1)^2 - 2)/1 = -1, so p = -1. The product of the new roots is (a/b)(b/a) = 1, so q = 1. Therefore, p + q = -1 + 1 = 2.

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