Question #2927a

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Answer 1 · 2017-04-29T06:32:44

#x=2 ; x=3#

The exponent is #2# so we will get two answers for #x# and that means two sets of brackets:

#x^2-5x+6=0#

#(x ....)(x ....)=0#

To obtain a #-5# in the middle term the factors of #+6# will need to be #2 and 3#, with both being negative:

#(x-2)(x-3)=0#

#(x-2)=0#

#x=2#

#(x-3)=0#

#x=3#

Substitute the values into the #given# equation to check:

#x^2-5x+6=0#

#2^2-5(2)+6=0#

#4-10+6=0#