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- February 5th 2014, 06:33 PM #1

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## x^3 - 6x^2 + 13x - 12 = 0

The method is than involving the formula (1/2)(-k+srt(k^2 +4h^3)) and the book is Liebeck's "A Concise Introduction to Pure Mathematics". Hopefully that is familiar to someone.

My attempt:

x^3 - 6x + 13x - 12 = 0

Let y = x-2

So y^3 = x^3 - 6x + 12x - 8

So y^3 + x - 4 = 0

So y^3 + y - 2 = 0

h = 1/3 , k = -2

Pluging into the formula gives y = crt((9+2srt(2))/9) + crt((9-2srt(2))/9)

From here I am totally stuck? Is this even right so far?

thanks

- February 5th 2014, 06:57 PM #2

- February 5th 2014, 06:59 PM #3

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## Re: x^3 - 6x^2 + 13x - 12 = 0

Hey kinhew93.

In terms of reconciling the roots, note that if 3 is a root then you can factor your polynomial by dividing by (x-3) and you should have no remainder. In other words:

P(x) = x^3 - 6x^2 + 13x - 12 = (x-3)(ax^2 + bx + c). You can use this to find a, b, and c and then use the quadratic formula to verify your other roots.

This is in addition to the cubic formula (which I haven't seen for a long time since first year university) and you can use the above to check your answer.

- February 5th 2014, 07:12 PM #4

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- February 5th 2014, 08:10 PM #5

- February 6th 2014, 01:07 AM #6

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- February 6th 2014, 02:59 AM #7

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## Re: x^3 - 6x^2 + 13x - 12 = 0

The specific method you are asked to use looks a lot like the equation numbered (20) on this page Cubic Formula -- from Wolfram MathWorld

The notation they use is a little different though

- February 6th 2014, 12:37 PM #8

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## Re: x^3 - 6x^2 + 13x - 12 = 0

The method is known as Vieta's method.

(Francois Viete (aka Franciscus Vieta) 1540-1603)

Anyone interested in the history and development of mathematics should find it interesting.

The first step, to eliminate the squared term, has been completed (successfully).

Next is to use a second substitution, y = w - 1/(3w) and that will get you a quadratic in w cubed.

Solve that using the usual formula for a quadratic and you have two values for w cubed.

Calculate w (from either value, it doesn't matter which, they give you the same value for y,) and back substitute for y and then for x.

The other two roots, which could be complex, can be calculated either by factoring out the solution just calculated and then solving the resulting quadratic, or by finding the two other (complex) values of w and then again back substituting twice.