How do you solve #e^ { 4} - 5e ^ { 2} + 4= 0#?

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Answer 1 · 2017-02-25T16:10:35

See the entire solution process below:

First, factor the right side of the equation playing with the factors of #4# (1x4, 2x2) to get:

#(e^2 - 4)(e^2 - 1) = 0#

Now, equate each term to #0# and solve for #e# to get the solutions.

Solution 1)

#e^2 - 4 = 0#

#e^2 - 4 + color(red)(4) = 0 + color(red)(4)#

#e^2 - 0 = 4#

#e^2 = 4#

#sqrt(e^2) = +-sqrt(4)#

#e = +-2#

Solution 2)

#e^2 - 1 = 0#

#e^2 - 1 + color(red)(1) = 0 + color(red)(1)#

#e^2 - 0 = 1#

#e^2 = 1#

#sqrt(e^2) = +-sqrt(1)#

#e = +-1#

The solution is #e = +-2# and #e = +-1# or #e = -2#, #e = 2#, #e = -1#, #e = 1#