help me with my maths please?

Question

1.Transform y=(3x+1)(x-2)+(x-3)squared in the vertex form...
2.Find the two positive integers whose sum is 19 & whose product is greater than or equal to 48 but less than 90. How many pairs of numbers will satisfy the conditions?
3.Find the value of k so that g(x)=kxsquared +3x -2 will have two real zeros.

...and please if you can, answer my first questions too. thanks!

Answer 1

y= 3x^2-6x+x-2 +x^2-6x+9 = 4x^2-11x+7
y=(2x-5.5)^2 -23.25 V(2.75,23.25)
a+b=19
48<=a*b<=90 b=19-a
so48 <19a-a^2 <90
a^2-19a+90 >0 a^2-19a+90= 0 a=((19+-sqrt(331-360)/2
so a<=5 or a =>9
for a >=9
a^2-19a+48<= 0
a= (19+-13)/ 2 so 3<=a<=16
sothe possible pairs are
(3,16) ok (4,15)ok (5,14)ok (9,10)ok (11,8) ,.(12,7) (13,6)
(14,5) the next (15,4) is the same as (4,15)
there are 8 pairs
3)9-8k>0 k<9/8