How do you solve #9s^{2} + 2= 12s#?

1 Answer
George C.
Sep 14, 2016

#s = 2/3+-sqrt(2)/3#

Explanation:

Subtract #12s# from both sides to get:

#0 = 9s^2-12s+2#

#color(white)(0) = 9s^2-12s+4-2#

#color(white)(0) = (3s)^2-2(2)(3s)+(2)^2-2#

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

#color(white)(0) = ((3s-2) - sqrt(2))((3s-2) + sqrt(2))#

#color(white)(0) = (3s-2 - sqrt(2))(3s-2 + sqrt(2))#

#color(white)(0) = (3s-(2+sqrt(2)))(3s-(2-sqrt(2)))#

Hence:

#s = (2+-sqrt(2))/3 = 2/3+-sqrt(2)/3#

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