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..

Guest Aug 27, 2014

#7**+10 **## The Laws

### Basic formulas

## Example

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

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 *x*_{1}, *x*_{2}, ..., *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.

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

#1**+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......)

CPhill Aug 27, 2014

#2**0 **

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$$$

Melody Aug 27, 2014

#3**+5 **

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.

Melody Aug 27, 2014

#4**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)

CPhill Aug 27, 2014

#5**+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

}$$

heureka Aug 28, 2014

#7**+10 **

Best Answer## The Laws

### Basic formulas

## Example

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

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 *x*_{1}, *x*_{2}, ..., *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.

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