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September 30, 2014

September 30, 2014

Posted by **Mary** on Sunday, January 3, 2010 at 1:57pm.

Solve over the complex numbers.

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

- Algebra 2 -
**Reiny**, Sunday, January 3, 2010 at 2:16pmI 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 -
**Mary**, Sunday, January 3, 2010 at 2:45pmThank you so much! I've been stuck on this problem for over 2 hours

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