Posted by Mary on Sunday, January 3, 2010 at 1:57pm.
"The Discriminant; Equations in Quadratic Form"
Solve over the complex numbers.
1/e + 1/(square root of e)  6 = 0

Algebra 2  Reiny, Sunday, January 3, 2010 at 2:16pm
I would let √e = x , then
1/x^2 + 1/x  6 = 0
multiply by x^2
1 + x  6x^2 = 0
6x^2  x  1 = 0
(3x+1)(2x1) = 0
x = 1/3 or x = 1/2
√e = 1/3 or √e = 1/2
e = 1/9 or e = 1/4
since we "squared" we have to verify all answers.
if e = 1/9
LS = 1/(1/9) + 1/√(1/9)  6
= 9 + 3  6 = 6
RS = 0, so e = 1/9 does not work
if e = 1/4
LS = 4 + 2  6 = 0 = RS
so e = 1/4
BTW, most texts and authors would avoid using e as a variable, since e is generally accepted as the Euler constant

Algebra 2  Mary, Sunday, January 3, 2010 at 2:45pm
Thank you so much! I've been stuck on this problem for over 2 hours
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