Very eerie result.

**Keithfert488** · March 21st 2010, 07:59 PM

Solve analytically for the intersections of

$y=4-\frac{x^4}{64}$

and

$y=\sqrt{16-x^2}$

I'm concerned solely with the intersection in the first quadrant. Please tell em your result. I want to know if I'm just crazy when I see mine. You will know what I'm talking about when you solve it.

**Wilmer** · March 21st 2010, 08:20 PM

x^4 = 256

and

x^2 = 16

**Keithfert488** · March 21st 2010, 08:30 PM

I'm talking about the one between 0 and 4

**pickslides** · March 21st 2010, 08:44 PM

Originally Posted by Keithfert488

I'm talking about the one between 0 and 4

I get $x= 0$

**Stroodle** · March 21st 2010, 08:47 PM

$x=2\sqrt{2\sqrt{5}-2}$

**Keithfert488** · March 21st 2010, 08:47 PM

I am talking about the solution where x is greater than 0 and less than 4

**Keithfert488** · March 21st 2010, 08:49 PM

Originally Posted by Stroodle

$x=2\sqrt{2\sqrt{5}-2}$

what I find so eerie is how close this is to pi. Is it just a coincidence?

**Stroodle** · March 21st 2010, 08:51 PM

Oh, I wouldn't have a clue

But I'd say it's probably just a coincidence...

**Keithfert488** · March 21st 2010, 08:56 PM

The reason I did this is because I was trying to solve for the intersection of a fourth root polynomial with y-intercept r and zeros -r and r and a circle centered at the origin with radius r. So the x-coordinate of that intersection is very close to $\frac{r\pi}{4}$

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