Question

If alpha and beta are zeros of f(x) = 2(x^2) - 5x + 7, find the polynomial whose zeros are (2 alpha) + (3 beta) and (3 alpha + 2 beta).

Answer 1

given:

f(x) = 2x^2 - 5x + 7

= 2x^2 - 7x + 2x + 7

= x(2x - 7) - 1(2x - 7)

= (x-1)(2x-7)

so,we have :-

alpha = 1 and beta = 7/2

according to question :-

one of its zero = 2alpha + 3beta

=2(1) + 3(7/)

=2 + 21/2

=4+21/2. [By L.C.M dinominator part comes 2 ]

=25/2

second of the polynomial = 3alpha + 2beta

= 3(1) + 2(7/2)

= 3 + 14/2

=3 + 7

=10

Solution:-

sum of new zeros =25/2 + 10

=25 + 20/2 [2 is come by take L.C.M]

=45/2

product of zeros = 25/2 × 10

= 125

by general formula :-

k[x^2 - [sum of zeros] x + produc of zeros] = 0 {where k is non - zero constant }

=k[x^2 - (45/2)x + (125)] = 0

=x^2 - 45/2x + 125

therefore ,required polynomial ,

x^2 - 45/2x + 125

Hope , its help you.......