Question

lf the zeroes ofx^2-kx+6are in the ratio3:2,find k.

Answer 1

let the zeroes be 3y and 2y
9y^2- 3yk +6 =0
9y^2-3yk =-6

4y^2-2yk=-6

9y^2-3yk=4y^2-2yk
9y^2-4y^2=3yk-2yk
5y^2= yk
5y =k

I know its wrong but just tried....
Putting the values,
9y^2-15y+6
9y^2-9y-6y+6
9y (y-1) - 6 (y-1)
(y-1)(9y-6)

Roots of y=1, 2/3 (y=1, 9y=6)
So,
For the first root,
The zeroes= 3,2
K=5

For the second root
Zeroes= 2, 4/3
K=10/3

Answer 2

let f(x)= x2-kx +6

let alpha and beta are the zeros of f(x)

alpha/beta=3/2
alpha=3beta/2
now the sun of zeros=alpha+beta=-k/1
3beta/2+beta=-k/1
5beta=-2k
again,product of zeros =alphabeta =6
(3beta/2)beta=6
beta*beta=4
beta=+-2

putting the value of beta in eq.1

-2k=5betak=-5/2beta
-5/2(+-2)

k=+-5