/sci/ - Science & Math » Thread #12072131

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Anonymous Tue 01 Sep 02:47:16 2020 No.12072131 View Reply Original Report

Quoted By: >>12072134 >>12072211 >>12072215

I've tried for so long to find the zeroes of functions of this format. With Desmos it gives me an answer but I have no fucking clue how to do it myself. Any ideas?

In particular:

-x2 + 5 + ?(x2 - 4) = 0

Anonymous

Anonymous Tue 01 Sep 2020 02:47:55 No.12072134 Report

Quoted By:

>>12072131

Shit x^2 not x2

Anonymous

Anonymous Tue 01 Sep 2020 03:25:45 No.12072211 Report

Quoted By:

>>12072131
Use a zero finding algorithm

Anonymous

Anonymous Tue 01 Sep 2020 03:28:33 No.12072215 Report

Quoted By: >>12072234 >>12072271

>>12072131
move the square root over
square both sides
solve with the quartic equation

Anonymous

Anonymous Tue 01 Sep 2020 03:41:33 No.12072234 Report

Quoted By: >>12072271 >>12072303

>>12072215
actually not even quartic it will be a biquadratic
$x^2-5=\sqrt{x^2-4} \\<br />x^4-10x^2+25=x^2-4 \\<br />x^4-11x^2+29=0 \\<br />x^2=\frac{11\pm \sqrt{11^2-4*29}}{2} \\ <br />x=\pm \sqrt{\frac{11\pm \sqrt{11^2-4*29}}{2}}$

Anonymous

Anonymous Tue 01 Sep 2020 04:09:41 No.12072271 Report

Quoted By:

>>12072234
>>12072215
obviously plug in $11\rightarrow 2c+1, 29\rightarrow c^2+p, x\rightarrow k$ to get your general case.

Anonymous

Anonymous Tue 01 Sep 2020 04:20:47 No.12072303 Report

Quoted By: >>12072309

>>12072234
Does this really work? Why not simply use rational zeroes and use synthetic division?

Anonymous

Anonymous Tue 01 Sep 2020 04:22:29 No.12072309 Report

Quoted By:

>>12072303
>does this really work
what part of it is unbelievable?
you're just solving the quadratic equation for z=x^2, and then taking the square root of that solution to get the solutions for x
>synthetic division
sure, that seems like it would work too

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