How do you solve #(a - 6) ( a - 2) = - 1#?

1 Answer
George C.
Jul 24, 2017

#a=4+-sqrt(3)#

Explanation:

Given:

#(a-6)(a-2)=-1#

Transpose and add #1# to both sides to get:

#0 = (a-6)(a-2)+1#

#color(white)(0) = a^2-8a+12+1#

#color(white)(0) = a^2-8a+16-3#

#color(white)(0) = (a-4)^2-(sqrt(3))^2#

#color(white)(0) = ((a-4)-sqrt(3))((a-4)+sqrt(3))#

#color(white)(0) = (a-4-sqrt(3))(a-4+sqrt(3))#

Hence:

#a=4+-sqrt(3)#

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