SOLUTION: find y so that x^2+(y+1)x+y+4=0 has only one solution. (I am not asking you to solve the quadratic in terms of y)

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Question 763177: find y so that x^2+(y+1)x+y+4=0 has only one solution. (I am not asking you to solve the quadratic in terms of y)
Answer by ramkikk66(644) About Me  (Show Source):
You can put this solution on YOUR website!
x%5E2+%2B+%28y%2B1%29%2Ax+%2B+y+%2B+4+=+0
If it has only ONE solution (root), the LHS must of the form (x+a)^2.
(x+a)^2 = x^2 + 2*a*x + a^2
Equating the above to the given equation for coefficient of x, and the constant term
(y+4) must be equal to(a^2), or a = sqrt(y+4)
The middle term is the equivalent of 2*a*x, i.e. y+%2B+1+=+2%2Aa+=+2%2Asqrt%28y%2B4%29
i.e.2*sqrt(y+4) = y+1
Squaring both sides:
4%2A%28y%2B4%29+=+%28y%2B1%29%5E2
4%2Ay+%2B+16+=+y%5E2+%2B+2%2Ay+%2B+1
y%5E2+-+2%2Ay+-+15+=+0
%28y+-+5%29%2A%28y+%2B+2%29+=+0
y = 5 or y = -2
Taking the positive solution of highlight%28y+=+5%29, the original equation becomes:
x^2 + 6x + 9 = 0 which has only one solution x = -3.
:)