# Algebra 2

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"The Discriminant; Equations in Quadratic Form"

Solve over the complex numbers.

1/e + 1/(square root of e) - 6 = 0

• Algebra 2 -

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)(2x-1) = 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 -

Thank you so much! I've been stuck on this problem for over 2 hours

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