Question

y²-9y=4 solve by competing square

Answer 1

Hi friend!!

Given polynomial,

y²-9y-4=0

y²-2(9/2)y+(9/2)²-(9/2)²-4=0

(y-9/2)²-(81/4)-4=0

(y-9/2)²=97/4

y-9/2=±√97/2

y=(9±√97)/2

I hope this will help you :D

Answer 2

= y2-9y-4 = 0

= y2-9y+(9/2)^2-(9/2)^2-4 = 0. (9 . 1/2 = 9/2)
(adding and subtracting the 9/2).

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

= (y-9/2)^2-81/4-4 = 0

= (y-9/2)^2 = 81/4+4

= (y-9/2)^2 = 97/4

= y-9/2 = +-√97/2

= y = 9/2+-√97/2

= therefore y = 9+√97/2

and , y = 9-√97/2.