SOLUTION: Solve algebraic expression {{{ a^6 -12a^4 + 48a^2 - 61 = 0 }}} and find the real roots.

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Question 1184166: Solve algebraic expression +a%5E6+-12a%5E4+%2B+48a%5E2+-+61+=+0+ and find the real roots.
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200) About Me  (Show Source):
You can put this solution on YOUR website!


+f%28a%29+=+a%5E6+-12a%5E4+%2B+48a%5E2+-+61+

Graphing the function will help us see if our results are reasonable. Here is a graph:

graph%28400%2C400%2C-3%2C3%2C-20%2C20%2Cx%5E6-12x%5E4%2B48x%5E2-61%29

Note the graph is even (symmetrical with respect to the y-axis, because every exponent in the polynomial is even.

There is no general method for solving a 6th degree polynomial equation.

By far the easiest way to find the APPROXIMATE real roots is by graphing the function on a graphing calculator and finding the zeros from the graph.

My TI-83 shows the real roots to be +/- 1.599296855 to 9 decimal places, which is consistent with the graph.

You can get that same result using Newton's Method. Look on the internet for a video on how to use Newton's Method on a TI83 or TI84 calculator.

I would consider doing that very worthwhile, since Newton's Method is a very powerful tool.

Finally, there is a way to find the exact real roots for this example.

Since the function is even, let y=a^2 and write the equation as

x^3-12x^2+48x-61=0

Then look again on the internet and find the formula for the general solution of a cubic equation and use it; then remember that you are solving for a=sqrt(x), not for x.

It is a very ugly formula, which I have never tried to use; and I have no inclination to try it on your example. So if you want to find the exact real zeros, I say have fun...!

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve algebraic cross%28expression%29   EQUATION   +a%5E6+-12a%5E4+%2B+48a%5E2+-+61+=+0+   and find the real roots.
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            It looks as a miracle,  but I will give a precise solution to the given equation. 


The starting equation is

    +a%5E6+-12a%5E4+%2B+48a%5E2+-+61 = 0+.       (1)



Recall the standard formula  (x-y)^3 = x^3 - 3x^2y + 3x*y^2 - y^3.



In the formula, put  x = a^2,  y = 4.  You will get

    a%5E6+-12a%5E4+%2B+48a%5E2+-+61 = a%5E6+-12a%5E4+%2B+48a%5E2+-+64 + 3.



Hence, the given equation is EQUIVALENT to

    %28a%5E2-4%29%5E3 = -3.      (2)


It implies that     


    a%5E2-4 = root%283%2C-3%29,

or

    a%5E2-4 = - root%283%2C3%29.    (3)



Hence,  

    a%5E2 = 4+-+root%283%2C3%29,

    a = +/- sqrt%284-root%283%2C3%29%29.     (4)


Approximate numerical values of the roots (4) are  +/- 1.599297.

Solved.