# College Algebra

I'm sorry I wrote the problem wrong previously.

"The equation x^2-9x-1=0 has 2 solutions A and B. Where A < B.
x^2-9x-1=0
(x-9)x=1
x^2=9x+1
(x-9/2)^2-85/4=0

Solutions x=1/2(9- sq root 85)
and x=1/2(9 + sq root 85)

I also have a second similar problem. 2x^2+17x+2=0
x(2x +17)+2=0
16/273(x+17/4)^2=1
2(x+17/4)^2-273/8=0
Solutions x=1/4(-17- sq root 273)
or x=1/4(sq root 273-17)

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2. 👎 0
3. 👁 132
1. x^2-9x-1 = 0
x^2-9x = 1
x^2 - 9x + 81/4 = 1 + 81/4
(x - 9/2)^2 = 85/4
x - 9/2 = ±√85/2
x = 9/2 ±√85/2
= 1/2 (9±√85)

2x^2+17x+2 = 0
2x^2+17x = -2
2(x^2+17/2 x) = -1
x^2 + 17/2 + 289/16 = -1 + 289/16
(x + 17/4)^2 = 273/16
x + 17/4 = ±√273/4
x = -17/4 ±√273/4
= 1/4 (-17±√273)

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2. 👎 0
posted by Steve

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