Trigonometry

Hi Melody,

François Viète (Latin: Franciscus Vieta; 1540 – 23 February 1603), Seigneur de la Bigotière, was a French mathematician whose work on new algebra was an important step towards modern algebra, due to its innovative use of letters as parameters in equations

The Laws

Basic formulas

Any general polynomial of degree n

(with the coefficients being real or complex numbers and a_n ≠ 0) is known by the fundamental theorem of algebra to have n (not necessarily distinct) complex roots x₁, x₂, ..., x_n. Vieta's formulas relate the polynomial's coefficients { a_k } to signed sums and products of its roots { x_i } as follows:

Equivalently stated, the (n − k)th coefficient a_n−k is related to a signed sum of all possible subproducts of roots, taken k-at-a-time:

for k = 1, 2, ..., n (where we wrote the indices i_k in increasing order to ensure each subproduct of roots is used exactly once).

The left hand sides of Vieta's formulas are the elementary symmetric functions of the roots.

Example

Vieta's formulas applied to quadratic and cubic polynomial:

For the second degree polynomial (quadratic) , roots of the equation satisfy

The first of these equations can be used to find the minimum (or maximum) of P. See second order polynomial.

For the cubic polynomial , roots of the equation satisfy

In reverse order, we have......

3)    (x - (5 + i)) (x  - (5 - i)) = ((x - 5) - i)) ((x -5) + i)) = (x-5)^2 - i^2 = x^2 -10x + 25 - (-1) = x^2 - 10x +26

2) (x - (3 + 7i))(x -  (3 - 7i)) = (((x-3) - 7i)((x-3) + 7i) = (x-3)^2 - 49i^2 = x^2 - 6x + 9  -49i^2 = x^2 - 6x + 9 + 49 = x^2 - 6x + 58

1) ( x  - (2 + √5)) ( x - (2 - √5)) = ((x - 2) - √5) ((x-2) + √5) = (x-2)^2 - 5 = x^2 - 4x + 4 - 5 = x^2 - 4x - 1

There is nothing to "solve"  .....just  to "expand"

(Thanks to Melody for pointing out my earlier error......)

Question 1

ok I am going to think about this as I type.

The solutions to the quadratic formula are 

$$\\x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\\\ x=\frac{-b}{2a}\pm \frac{\sqrt{b^2-4ac}}{2a}\\\\\\$$

$$Now if the roots are $2\pm \sqrt5$ then\\\\ $\frac{-b}{2a}=2 \rightarrow b=-4a$\\\\and\\\\ $\frac{\sqrt{b^2-4ac}}{2a}=\sqrt5$\\\\ $\frac{\sqrt{b^2-4ac}}{\sqrt{(2a)^2}}=\sqrt5$\\\\ $\frac{b^2-4ac}{4a^2}=5$\\\\ $b^2-4ac=20a^2$\\\\ $16a^2-4ac=20a^2$\\\\ $a^2-4ac=4a^2$\\\\ $-4ac=4a^2$\\\\$$

$$$-c=a$\\ $c=-a$\\ so the coefficients are a, -4a and -a\\ $y=ax^2-4ax-a$\\ $y=a(x^2-4x-1) $ Where a is a constant and $a\ne 0$$ $

What have you answered Chris     These are quadratics.

Oh I just worked out what you did.  I sure found the hard way. 

I am not sure whether I should die of embarasment or die laughing.   Maybe a little of both!

I still don't know why you joined them all together though.

Thanks, Melody......I'm sorry I didn't read the question  closely enough.....I'll go back and provide an edit!!!

Thanks, again....(it's still too early for me, I guess)

Write the quadratic function whose zeros are:

$$\boxed{ &x^2+px+q=0\quad \begin{array}{rcll} -p&=&x_1+x_2&\mbox{[Vieta]}\\ q&=&x_1x_2&\mbox{[Vieta]} \end{array} }$$

1. 2+square root of 5, 2-square root of 5

$$\big{ \left\begin{array}{r*{3}{cl}} -p&=&(2+\sqrt{5})&+&(2-\sqrt{5})& =&4\\ q&=&(2+\sqrt{5})&*&(2-\sqrt{5}) &=&4-5=-1 \end{array} \right\} x^2-4x-1=0 }$$

2. 3+7i,3-7i

$$\big{ \left\begin{array}{r*{3}{cl}} -p&=&(3+7i )&+&(3-7i )& =&6\\ q&=&(3+7i )&*&(3-7i ) &=&9-49i^2=58 \end{array} \right\} x^2-6x+58=0 }$$

3. 5±i

$$\big{ \left\begin{array}{r*{3}{cl}} -p&=&(5+i )&+&(5-i)& =&10\\ q&=&(5+i)&*&(5-i) &=&25-i^2=26 \end{array} \right\} x^2-10x+26=0 }$$

Thank you Heureka    

What is Vieta 

Vieta's formulas are formulas that relate the coefficients of a polynomial to sums and products of its roots ...

http://en.wikipedia.org/wiki/Vieta's_formulas

Very cool Heureka!

Thanks Heureka.    

I will check it out   

Very nice, heureka.......I have not heard of this before........this is an area definitely worthy of a "new" exploration for me....!!!!