# What is a if x^4+x^3+ax^2+x+a=0 and s=-2?s=1/p-1+1/q-1+1/m-1+1/n-1

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The request of the problem is vague since it does not specify what are p,q,m,n, hence, considering p,q,m and n as values of roots of the given equation, you may use Vieta's relations such that:

`p+m+n+q = -1/1 = -1`

`pm + pn + pq + mn + mq + np = a/1 = a`

`pmn + pmq + pnq + mnq = -1/1 = -1`

`pmnq = a/1 = a`

Notice that the problem provides the information that `s=-2` and `s = 1/(p-1)+ 1/(q-1) + 1/(m-1) + 1/(n-1), ` hence `1/(p-1) + 1/(q-1) + 1/(m-1) + 1/(n-1)= -2.` Bringing the fractions to a common denominator yields:

`((q-1)(m-1)(n-1) + (p-1)(m-1)(n-1) + (p-1)(q-1)(n-1) + (p-1)(q-1)(m-1))/((p-1)(q-1)(m-1)(n-1)) = -2`

Opening the brackets yields:

`((qm-q-m+1)(n-1) + (pm-p-m+1)(n-1) + (pq-p-q+1)(n-1) + (pq-p-q+1)(m-1))/((pq-p-q+1)(mn-m-n+1))`

`(qmn - qm - mn + n - qm + q + m - 1 + pmn - pm - pn + p - mn + m + n - 1 + pqn - pq - pn + p - qn + q + n - n + pqm - pq - pm + p - qm + q + m - 1)/(pqmn - pqm - pqn + pq - pmn + pm + pn - p - qmn + qm + qn - q + mn - m - n + 1) = -2`

`(-1 - 2qm - 2mn - 2pm - 2pn -2pq- 2qn + 3(m+n+p+q) - 3)/(a + 1 + a + 2) = -2`

`(-1 - 2a - 3 - 3)/(2a+3) = -2 => -2a - 7 = 2a + 3`

`4a = 10 => a = 10/4 => a = 5/2`

**Hence, evaluating a considering m,n,p,q as solutions to the given equation, yields `a = 5/2` .**