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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
Guest Aug 27, 2014

Best Answer 

 #7
avatar+20024 
+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

P(x)=a_nx^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0 \,

(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:

\begin{cases} x_1 + x_2 + \dots + x_{n-1} + x_n = -\tfrac{a_{n-1}}{a_n} \\ (x_1 x_2 + x_1 x_3+\cdots + x_1x_n) + (x_2x_3+x_2x_4+\cdots + x_2x_n)+\cdots + x_{n-1}x_n = \frac{a_{n-2}}{a_n} \\ {} \quad \vdots \\ x_1 x_2 \dots x_n = (-1)^n \tfrac{a_0}{a_n}. \end{cases}

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:

\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} x_{i_1}x_{i_2}\cdots x_{i_k}=(-1)^k\frac{a_{n-k}}{a_n}

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) P(x)=ax^2 + bx + c, roots x_1, x_2 of the equation P(x)=0 satisfy

 x_1 + x_2 = - \frac{b}{a}, \quad x_1 x_2 = \frac{c}{a}.

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

For the cubic polynomial P(x)=ax^3 + bx^2 + cx + d, roots x_1, x_2, x_3 of the equation P(x)=0 satisfy

 x_1 + x_2 + x_3 = - \frac{b}{a}, \quad x_1 x_2 + x_1 x_3 + x_2 x_3 = \frac{c}{a}, \quad x_1 x_2 x_3 = - \frac{d}{a}.
heureka  Aug 28, 2014
 #1
avatar+89889 
+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
avatar+93654 
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
avatar+93654 
+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
avatar+89889 
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
avatar+20024 
+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
 #6
avatar+93654 
0

Thank you Heureka    

What is Vieta 

Melody  Aug 28, 2014
 #7
avatar+20024 
+10
Best Answer

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

P(x)=a_nx^n + a_{n-1}x^{n-1} +\cdots + a_1 x+ a_0 \,

(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:

\begin{cases} x_1 + x_2 + \dots + x_{n-1} + x_n = -\tfrac{a_{n-1}}{a_n} \\ (x_1 x_2 + x_1 x_3+\cdots + x_1x_n) + (x_2x_3+x_2x_4+\cdots + x_2x_n)+\cdots + x_{n-1}x_n = \frac{a_{n-2}}{a_n} \\ {} \quad \vdots \\ x_1 x_2 \dots x_n = (-1)^n \tfrac{a_0}{a_n}. \end{cases}

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:

\sum_{1\le i_1 < i_2 < \cdots < i_k\le n} x_{i_1}x_{i_2}\cdots x_{i_k}=(-1)^k\frac{a_{n-k}}{a_n}

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) P(x)=ax^2 + bx + c, roots x_1, x_2 of the equation P(x)=0 satisfy

 x_1 + x_2 = - \frac{b}{a}, \quad x_1 x_2 = \frac{c}{a}.

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

For the cubic polynomial P(x)=ax^3 + bx^2 + cx + d, roots x_1, x_2, x_3 of the equation P(x)=0 satisfy

 x_1 + x_2 + x_3 = - \frac{b}{a}, \quad x_1 x_2 + x_1 x_3 + x_2 x_3 = \frac{c}{a}, \quad x_1 x_2 x_3 = - \frac{d}{a}.
heureka  Aug 28, 2014
 #8
avatar
+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
Guest Aug 28, 2014
 #9
avatar+93654 
0

Thanks Heureka.    

I will check it out   

Melody  Aug 30, 2014
 #10
avatar+89889 
0

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

 

CPhill  Aug 30, 2014

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