How do you solve #(2s-1)^{2}=225#?

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Answer 1 · 2017-06-18T06:07:32

#x=-7# or #x=8#

There is a rule that states #(a+b)^2=a^2+2ab+b^2#

In this case, #a# is #2s# and #b# is #-1#.

Therefore

#4s^2-4s+1=225#

Subtract #225# from both sides.

#4s^2-4s-224=0#

Divide by #4#.

#s^2-s-56=0#

You can factorise this as follows :

#(x+7)(x-8)#

Set each factor equal to #0#.

#x+7=0#
#x=-7#

#x-8=0#
#x=8#

Answer 2 · 2017-06-18T07:26:30

#s = 8 or s = -7#

Although this equation leads to a quadratic trinomial, it is in the form #x^2 =c" "# so we can just find the square root of both sides,

#(2s-1)^2 = 225#

#2s-1 = +-sqrt225#

#2s = +-sqrt225+1#

#s = (+15+1)/2 = 8#

#s = (-15+1)/2 = -7#