Question

if alpha and beta are the zeroes of the quadratic polynomial:p(x)-3x^2-4x+1,find a quadratic polynomial whose zeroes are alpha^2/beta and beta^2/alpha.​

Answer 1

Α and β are the zeros of 3x²-4x +1 polynomial,

first of all we factorise 3x²-4x+1

3x² -4x + 1

=3x² -3x -x +1

=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

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

= 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

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Answer 2

For the quadratic equation with roots, α ,β

Sum of roots = 4/3

Product of roots = 1/3 .

Now, The quadratic equation with roots α²/β ,

β²/ α

Now,

Sum of roots = α²/β + β²/ α = α³+β³/αβ

= (α + β)³-3αβ(α+β)/αβ

= (4/3)³-3(1/3)(4/3) / 1/3

= 64/27 - 4/3 / 1/3

= 64/27-36/27 / 1/3

= 28/27 * 3/1

= 28/9

Product of roots = (α²/β * β²/ α) = (αβ ) = 1/3 .

The quadratic equation with required roots = x²-28/9x + 1/3 = 9(x² - 28/8x+1/3) = 9x² - 28x + 3