Question

can anyone of u give me the correct answer

if u don't know.....dont spam ​

Answer 1

Step-by-step explanation:

Given equation is

$3 {x}^{2} - 6x + 1 = 0$

$its \: \: zeros \: \: are \: \: \alpha \: \: and \: \: \beta$

So, sum of zeros (α+β)= -b/a=2 and product of zeros=c/a=1/3

A polynomial whose zeros are α²β & β²α is

$k( {x}^{2} - ( { \alpha }^{2} \beta + { \beta }^{2} \alpha )x + ({ \alpha }^{2} \beta) .( { \beta }^{2} \alpha) )$

,where k is a constant and belongs to R

$= > k( {x}^{2} - \alpha \beta ( \alpha + \beta )x + ({ \alpha \beta })^{3} )$

Now putting the values, we get,

$= > k( {x}^{2} - \frac{2}{3} x + \frac{1}{27} )$

Answer 2

Step-by-step explanation:

$\huge\underline\mathfrak\color{red} Answer$

$α \: and \: β \: are \: the \: zeros \: of \: 3x²-4x +1 \: polynomial, \: </p><p></p><p>$

first of all we factorise :-

$3x²-4x+1</p><p>3x² -4x + 1</p><p>$

$=3x² -3x -x +1</p><p></p><p>$

$=3x( x -1) -1(x -1)$

$=(3x -1)(x -1)$

hence. (3x -1) and (x -1) are the factors of given polynomial .

so, x = 1/3 and 1 are the zeros of that polynomial.

hence, α = 1/3. and β = 1

$=(3x -1)(x -1)$

or α =

$1 and β. = 1/3$

you can choose any one in both

I choose α =

$1. and β = 1/3$

now,

let any unknown. polynomial. whose zeros are

$α²/β and β²/α$

$α²/β = (1)²/(1/3) = 3$

$β²/α = (1/3)²/1 = 1/9$

now, equation of unknown polynomial.

x²- ( sum of roots)x + product of roots

$β²/α = (1/3)²/1 = 1/9$

$= x²- ( α²/β + β²/α)x +(α²/β)(β²/α)$

$put α²/β = 3 and β²/α = 1/9$

$= x²- ( 3 +1/9)x + 3 × 1/9$

$= x² -28x/9 + 3/9$

$={ 9x² -28x + 3 }1/9$

hence, 9x² -28x + 3 is answer;!

HOPE U GOT IT NOW ADEELA!

<MY BROTHER HELPED ME:D>

it took 1 hour finding this!

nd if u can then plz mrk as brainliest answer as well adeela didi ❤✌