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

2 Answers
Jun 18, 2017

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

Explanation:

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#

Jun 18, 2017

#s = 8 or s = -7#

Explanation:

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#