How do you solve #5(x-7)^2=135#?

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Answer 1 · 2017-06-05T07:16:14

12.2 & 1.8

Given, #5(x-7)^2=135#

#rArr x^2-14x+49=135/5#

#rArr x^2-14x+49-27=0#

#rArr x^2-14x+22=0#

#rArr x = [-b+-sqrt{b^2-4ac}]/[2a]# [here b = -14, c=22 & a = 1]

#rArr x = [-(-14)+-sqrt{(-14)^2-4*1*22}]/[2*1]#

#rArr x = [14+-sqrt(196-88)]/2#

#rArr x = [14 +-sqrt108]/2#

#rArr x = (14+-10.4)/2#

#rArr x = [14+10.4]/2 , [14-10.4]/2#

#rArr x = 12.2, 1.8#

Answer 2 · 2017-06-05T07:30:56

#x =12.196 or x = 1.804#

Isolate the bracket that contains the #x# by dividing both sides by #5#

#(5(-7)^2)/5 = 135/5#

#(x-7)^2 = 27#

Find the square root of both sides:

#x-7 = +-sqrt27#

Solve using the positive and negative roots:

#x = +sqrt27 +7 = 12.196#

#x = -sqrt27+7 = 1.804#