How do you solve #y^2-5y-3=0# by completing the square?

2 Answers
Apr 29, 2016

Answer:

#y = 5/2+-sqrt(37)/2#

Explanation:

We will also use the difference of squares identity, which can be written:

#a^2-b^2=(a-b)(a+b)#

with #a=(2y-5)# and #b=sqrt(37)#.

Pre-multiply by #4# to reduce the amount of working in fractions:

#0 = 4(y^2-5y-3)#

#=4y^2-20y-12#

#=(2y-5)^2-25-12#

#=(2y-5)^2-(sqrt(37))^2#

#=((2y-5)-sqrt(37))((2y-5)+sqrt(37))#

#=(2y-5-sqrt(37))(2y-5+sqrt(37))#

Hence:

#y = 5/2+-sqrt(37)/2#

May 2, 2016

Answer:

#y = 5.541# OR # y = -0.541#

Explanation:

#y^2 - 5y = 3 rArr# move the constant to the right hand side.

Complete the square by adding what is missing from the square of the binomial to BOTH sides. #(b÷2)^2#

#y^2 - 5y + color(red) (5/2)^2 = 3 + color(red)(5/2)^2#

#(y - 5/2)^2 = 3 + (25/4)#

#y - 5/2 = +-sqrt(9.25# ............#rArr 3+6 1/4 = 9 1/4#

#y = sqrt9.25 +2.5# OR # y = -sqrt9.25 +2.5#

#y = 5.541# OR # y = -0.541#