Question

16x^2-72+81y^2-12x+27y+9/4​

Answer 1

Step-by-step explanation:

16x^2-72xy+81y^2-12x+27y+9/4=0............(1),

take 4x^2-72xy+81y^2 = (4x-9y)^2

let the two factors are 4x-9y+l=0 and 4x-9y+m=0

multiply the two factor to get equation (1)

(4x-9y+l)(4x-9y+m)=0 (on multiplication we get three terms are same as in (1)

16x^2-72xy+81y^2+(4l+4m)x+(-9l-9m)y+lm=0.............(2)

(1) and (2) represent same equation, equate the coefficients

4l+4m=-12, -9l-9m=27 => l+m = -3, and lm = 9/4 => l = 9/4m

substitute l=9/4m in l+m = -3

(9/4m)+m = -3

9+4m^2 = -12m

4m^2+12m+9=0

(2m+3)^2=0

2m+3=0

m = -3/2

If m=-3/2 l = 9/[4.(-3/2)] = 9/(-6) => l = - 3/2

thus the two factors are

4x-9y-3/2=0 and 4x-9y-3/2=0 that is

(4x-9y-3/2)^2=0