If one zero of f(x)=4x2-8kx+8x-9 is negative of the . . .

Question : If one zero of f(x)=4x²-8kx+8x-9 is negative of the other, then find zeroes of kx²+3kx+2.

Doubt by Angel

Solution : 

f(x)=4x²-8kx+8x-9
f(x)=4x²-8(k-1)x-9
Here 

a=4
b=-8(k-1)
c=-9

Let 

α=α
β=-α

Sum of zeroes
α+β=-b/a
α+(-α)=-[-8(k-1)]/4
0=8(k-1)/4
0=2(k-1)
0/2=k-1

0=k-1
1=k
k=1 — (1)

=kx²+3kx+2
=(1)x²+3(1)x+2 [From equation (1)]
=x²+3x+2
=x²+(2+1)x+2
=x²+2x+x+2
=x(x+2)+1(x+2)
=(x+1)(x+2)

x+1=0
x=-1

x+2=0
x=-2

Hence, required zeroes of kx²+3kx+2 are -1 and -2.