How do you solve #6p^2-17p+12=0# by completing the square?

1 Answer
Apr 29, 2018

#p=3/2 or p=4/3#
To Complete the square use the formula given below.You can check it for different squares,It is always TRUE.
#color(blue)((psi) to III^(rd)term=(II^(nd)term)^2/(4xxI^(st)term)#

Explanation:

Here,

#6p^2-17p+12=0...to(1)#

We find third term #X#,such that

#(6p^2-17p+X) is # square and then find #y=12-X#

#I^(st)term=6p^2,#

#II^(nd)term=-17p,#

# III^(rd)term=X#

Note:

#color(blue)((psi) to III^(rd)term=(II^(nd)term)^2/(4xxI^(st)term)= (-17p)^2/(4xx(6p^2))=289/24#

#i.e. X=289/24=>y=12-289/24=(288-289)/24=-1/24#

From #(1)#

#6p^2-17p+289/24-1/24=0#

#=>(sqrt6p-17/(2sqrt6))^2=1/24=(1/(2sqrt6))^2#

#=>sqrt6p-17/(2sqrt6)=+-1/(2sqrt6)#

#=>sqrt6p-17/(2sqrt6)=+1/(2sqrt6)orsqrt6p- 17/(2sqrt6)=-1/(2sqrt6)#

#=>sqrt6p=17/(2sqrt6)+1/(2sqrt6) or sqrt6p=17/(2sqrt6)-1/(2sqrt6)#

#sqrt6p=18/(2sqrt6) or sqrt6p=16/(2sqrt6)#

#p=18/(2xx6) orp=16/(2xx6)#

#p=3/2 or p=4/3#