How do you solve $3tan^2x+5tan x-1=0$ ?

Question

5471 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-10-22T07:47:51

$x = tan^(-1)(-5/6+-sqrt(37)/6) + npi" "$ for any integer $n$

Given:

$3tan^2x+5tan x-1=0$

Let:

$t = tan x$

Then our equation becomes:

$3t^2+5t-1=0$

This is in the form:

$at^2+bt+c = 0$

which has discriminant $Delta$ given by the formula:

$Delta = b^2-4ac = 5^2-4(3)(-1) = 25+12 = 37$

Since this is positive, the quadratic equation in $t$ has real roots, but since it is not a perfect square those roots are irrational.

We can use the quadratic formula to find:

$t = (-b+-sqrt(b^2-4ac))/(2a)$

$color(white)(t) = (-b+-sqrt(Delta))/(2a)$

$color(white)(t) = (-5+-sqrt(37))/6$

That is:

$tan x = -5/6+-sqrt(37)/6$

Note that $tan x$ has period $pi$ .

So:

$x = tan^(-1)(-5/6+-sqrt(37)/6) + npi" "$ for any integer $n$

How do you solve 3tan^2x+5tan x-1=03tan^2x+5tan x-1=0 ?