How do you solve #2tan^2(t)+ 8tan(t)+ 4 = 0#?

1 Answer
Nghi N.
Aug 22, 2016

#106^@34 + k180#
#149^@50 + k180^@#

Explanation:

Call tan t = x and solve the below quadratic equation for x:
#2x^2 + 8x + 4 = 0#
Use the improved quadratic formula (Socratic Search):
#D = d^2 = b^2 - 4ac - 64 - 32 = 32# --> #d = +- 4sqrt2#
There are 2 real roots:
#tan t = x = -b/(2a) +- d/(2a) = -8/4 +- (4sqrt2)/4 = -2 +- sqrt2#

a. #tan t = -2 + sqrt2 = - 2 + 1.41 = - 0.59#
Calculator -->
#t = - 30^@54#, or #x = 149^@50# (co-terminal)
b. #tan t = - 2 - sqrt2 = - 3.41#.
Calculator gives:
#t = - 73^@66# or #t = 106^@34# (co-terminal)
General answers:
#106^@34 + k180^@#
#149^@50 + k180^@#

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