Question

How do you use the quadratic formula to solve `tan^2x+3tanx+1=0` in the interval `[0,2pi)`?

Answer 1

`20^@90, 84^@86, 264^@86, 339^@10`

Explanation:

Solve this quadratic equation for tan x, by using the improved quadratic formula (Socratic Search):
`tan^2 x + 3tan x + 1 = 0`
`D = d^2 = b^2 - 4ac = 9 - 4 = 5` --> `d = +- sqrt5`
There are 2 real roots:
`tan x = -b/(2a) +- d/(2a) = -3/2 +- sqrt5/2`
`tan x1 = (-3 + sqrt5)/2 = - 0.76/2 = - 0.38`
`tan x2 = (-3 - sqrt5)/2 = 22.26/2 = 11.12`
Use calculator and unit circle -->
tan x1 = - 0.38 --> `x1 = - 20^@90` and
`x1 = - 20.90 + 180 = 159^@90`
The arc `(-20^@90)` is co-terminal to arc:`(360 - 20.90 = 339^@0)`
tan x2 = 11.12 --> `x2 = 84^@86` and
`x2 = 84.86 + 180 = 264^@86`
Answers for `(0, 360)`:
`20^@90, 84^@86, 264^@86, 339^@10`