Question

(1) If alpha and beta are the zeroes of the polynomaial 2X(square)-5X+7 , Find a polynomial whose zeroes are 2 alpha + 3 beta and 3 alpha + 2 beta.

(2) If alpha and beta are the zeroes of the polynimial 3X(square)-4X+1 , Find a polynomial whose zeroes are alpha(square) by (divided by ) beta and beta (square) by alpha.

please help me....

Answer 1

α² + β² can be written as (α + β)² - 2αβ

p(x) = 2x² - 5x + 7
a = 2 , b = - 5 , c = 7

α and β are the zeros of p(x)

we know that ,
sum of zeros = α + β
= -b/a
= 5/2

product of zeros = c/a
= 7/2


2α + 3β and 3α + 2β are zeros of a polynomial.

sum of zeros = 2α + 3β+ 3α + 2β
= 5α + 5β
= 5 [ α + β]
= 5 × 5/2
= 25/2

product of zeros = (2α + 3β)(3α + 2β)
= 2α [ 3α + 2β] + 3β [3α + 2β]
= 6α² + 4αβ + 9αβ + 6β²
= 6α² + 13αβ + 6β²
= 6 [ α² + β² ] + 13αβ
= 6 [ (α + β)² - 2αβ ] + 13αβ
= 6 [ ( 5/2)² - 2 × 7/2 ] + 13× 7/2
= 6 [ 25/4 - 7 ] + 91/2
= 6 [ 25/4 - 28/4 ] + 91/2
= 6 [ -3/4 ] + 91/2
= -18/4 + 91/2
= -9/2 + 91/2
= 82/2
= 41

-18/4 = -9/2 [ simplest form ]

a quadratic polynomial is given by :-

k { x² - (sum of zeros)x + (product of zeros) }

k {x² - 5/2x + 41}

k = 2

2 {x² - 5/2x + 41 ]

2x² - 5x + 82 -----> is the required polynomial



2.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
= 28/24 * 3/1
= 28/8
= 14/4
= 7/2 .

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

The quadratic equation with required roots = x²-7/2x + 1/3 = 6x² - 21x + 2