How do you solve #y^2-6y+8=0#?

3 Answers
Alan P.
Mar 14, 2016

Factor the left side to find
#color(white)("XXX")y=2 or y=4#

Explanation:

Given
#color(white)("XXX")y^2-6y+8=0#

Factoring the left side expression
#color(white)("XXX")(y-4)(y-2)=0#

which implies
either
#color(white)("XXX")(y-4)=0 rarr y=4#
or
#color(white)("XXX")(y-2)=0 rarr y=2#

ali ergin
Mar 14, 2016

#{2,4}#

Explanation:

#y^2-6y+8=0#
#(y-2)(y-4)=0#
#(y-2)=0 => y=2#
#(y-4)=0 =>y=4#
#{2,4}#

Andy Y.
Mar 14, 2016

#y=2# or #y=4#

Explanation:

Method of factoring:

#y^2-6y+8=0#
#(y-2)(y-4)=0#
#y-2=0# or #y-4=0#

Solving for #y#, gives #y=2# or #y=4#

Method of quadratic equation:

#y=(6+-sqrt(36-4(1)(8)))/(2(1))=(6+-sqrt(4))/(2)#
#y=2# or #y=4#

Method of graphing:
graph{x^2-6x+8 [-0.927, 9.073, -1.44, 3.56]}

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