# How to turn this periodic function in a quadratic equation?

• Mar 6th 2010, 04:24 AM
eriiin
How to turn this periodic function in a quadratic equation?
I have acquired the equation:

0 = 70tanθ - 1 / 15.3125 (1+tan² 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, 04:51 AM
Prove It
Quote:

Originally Posted by eriiin
I have acquired the equation:

0 = 70tanθ - 1 / 15.3125 (1+tan² 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 (:

I can't read this. Please either use LaTeX or brackets where they're needed.

Is it $0 = 70\tan{\theta} - \frac{1}{15.3125(1 + \tan^2{\theta})}$?
• Mar 6th 2010, 04:53 AM
eriiin
Yes sorry, that is correct.
• Mar 6th 2010, 05:02 AM
Prove It
Start by getting a common denominator:

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

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

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

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

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

You may need technology to solve this equation.
• Mar 6th 2010, 06:08 AM
eriiin
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, 07:36 AM
wintersolstice
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, 07:40 AM
eriiin
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, 08:07 AM
wintersolstice
Quote:

Originally Posted by eriiin
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.

If you put:

a = 1071.875 ( $tan^3$)

b = 0 ( $tan^2$)

c = 1071.875 ( $tan$)

d = -1

so it can be solved the " $tan^2$" doesn't appear but it exists (this is true in all polynomials)

I hope that explains it