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# Trigonometry

0
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Write the quadratic function whose zeros are:

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

2. 3+7i,3-7i

3. 5±i

Please tell me how to solve them... I don't get the question..

math algebra
Aug 27, 2014

#7
+20850
+10

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 an ≠ 0) is known by the fundamental theorem of algebra to have n (not necessarily distinct) complex roots x1x2, ..., xn. Vieta's formulas relate the polynomial's coefficients { ak } to signed sums and products of its roots { xi } as follows:

Equivalently stated, the (n − k)th coefficient ank 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 ik 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

Aug 28, 2014

#1
+94618
+10

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......)

Aug 27, 2014
#2
+95361
0

Question 1

The solutions to the quadratic formula are

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

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

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

.
Aug 27, 2014
#3
+95361
+5

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.

Aug 27, 2014
#4
+94618
0

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)

Aug 27, 2014
#5
+20850
+5

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

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

.
Aug 28, 2014
#6
+95361
0

Thank you Heureka

What is Vieta

Aug 28, 2014
#7
+20850
+10

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 an ≠ 0) is known by the fundamental theorem of algebra to have n (not necessarily distinct) complex roots x1x2, ..., xn. Vieta's formulas relate the polynomial's coefficients { ak } to signed sums and products of its roots { xi } as follows:

Equivalently stated, the (n − k)th coefficient ank 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 ik 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

heureka Aug 28, 2014
#8
+5

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

Heureka

Very cool Heureka!

Heureka
heureka
heureka
Aug 28, 2014
#9
+95361
0

Thanks Heureka.

I will check it out

Aug 30, 2014
#10
+94618
0

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

Aug 30, 2014