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

1. ## 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 (:

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

3. Yes sorry, that is correct.

4. 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.

5. 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:

.

to a tan^2.

Any guidance to how I do this?

6. 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

7. 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.

8. 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