# Determine the set of values of m for which both roots of equation x^2 - (m+1)x + m+4 = 0 are real and negative.

## Expert Answers x^2 - (m+1)x + m + 4 = 0

If the equation has ral roots, then the discriminant must be positibs:

Let us verify:

d = b^2 - 4ac > 0

==> (m+1)^2 - 4*1* (m+4) > 0

==> m^2 + 2m + 1 - 4m - 16 > 0

==> m^2 -2m - 15  > 0

==> (m-5)(m+3) >0

==> m-5 > 0   and (m+3)>0   OR (m-5)< 0  and m+3 < 0

==> m > 5  and m > -3            OR    m < 5    and m < -3

==> m belongs to ( 5, inf) U (-inf, -3)...........(1)

Now in order for the roots to be negative then :

x1+ x2 < 0    and x1*x2 > 0

But x1 + x2 = -b/a = (m+1) < 0 ==> m < -1`

Also, x1*x2 = c/a = (m+4) > 0 ==> m  > -4

==> m belongs to ( -inf, -1) U (-4, inf)...........(2)

From (1) and (2) we conclude:

m belongs to  (-inf, -3) U (5, inf )

Approved by eNotes Editorial Team

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