I have acquired the equation:

0 = 70tanθ - 1 / 15.3125 (1+tan²θ)

I understand that I can solve this by turning it into a quadratic, just unsure how you rearrange this so that it is a quadratic.

Any help would be extremely appreciated (:

Printable View

- Mar 6th 2010, 03:24 AMeriiinHow to turn this periodic function in a quadratic equation?
I have acquired the equation:

0 = 70tanθ - 1 / 15.3125 (1+tan**²**θ)

I understand that I can solve this by turning it into a quadratic, just unsure how you rearrange this so that it is a quadratic.

Any help would be extremely appreciated (: - Mar 6th 2010, 03:51 AMProve It
- Mar 6th 2010, 03:53 AMeriiin
Yes sorry, that is correct.

- Mar 6th 2010, 04:02 AMProve It
Start by getting a common denominator:

$\displaystyle 0 = 70\tan{\theta} - \frac{1}{15.3125(1 + \tan^2{\theta})}$

$\displaystyle 0 = \frac{1071.875\tan{\theta}(1 + \tan^2{\theta})}{15.3125(1 + \tan^2{\theta})} - \frac{1}{15.3125(1 + \tan^2{\theta})}$

$\displaystyle 0 = \frac{1071.875\tan{\theta}(1 + \tan^2{\theta}) - 1}{15.3125(1 + \tan^2{\theta})}$

$\displaystyle 0 = 1071.875\tan{\theta}(1 + \tan^2{\theta}) - 1$

$\displaystyle 0 = 1071.875\tan^3{\theta} + 1071.875\tan{\theta} - 1$.

You may need technology to solve this equation. - Mar 6th 2010, 05:08 AMeriiin
I am unsure to whether I am doing this correctly, because I've never done a quadratic that has a number with a cube root before.

So I am guessing I need to somehow get the:

http://www.mathhelpforum.com/math-he...22b83028-1.gif.

to a tan^2.

Any guidance to how I do this? - Mar 6th 2010, 06:36 AMwintersolstice
Perhaps you could solve it as a "cubic" instead of a "quadratic"

here's the cubic formula:

Cubic function - Wikipedia, the free encyclopedia

hope that helps - Mar 6th 2010, 06:40 AMeriiin
Never even seen that equation before /:

But I only have three numbers which I can use, not four, so I take it that idea isn't possible. - Mar 6th 2010, 07:07 AMwintersolstice