How do you solve #(x-2)^2-25=0#?

3 Answers
Aug 24, 2016

Answer:

#x=7#

#x=-3#

Explanation:

#(x-2)^2-25=0#

or

#(x-2)^2=25#

or

#x-2=sqrt25#

or

#x-2=+-5#

or

#x=2+-5#

or

#x=2+5#

or

#x=7#---------------------------------Ans #1#

or

#x=2-5#

or

#x=-3#-----------------------------Ans #2#

Aug 24, 2016

Answer:

#x=7,x=-3#

Explanation:

First step is to isolate the #(x-2)^2# term

add 25 to both sides.

#rArr(x-2)^2-cancel(25)+cancel(25)=0+25#

#rArr(x-2)^2=25#

now take the #color(blue)"square root of both sides"#

#sqrt((x-2)^2)=+-sqrt25#

#rArrx-2=+-5#

solve : #x-2=5rArrx=5+2=7#

solve : #x-2=-5rArrx=-5+2=-3#

Aug 24, 2016

Answer:

#x=7# or #x=-3#

Explanation:

We can use identity #(a^2-b^2)=(a-b)(a+b)#. As such

#(x-2)^2-25=0#

#hArr(x-2)^2-5^2=0#

or #(x-2-5)(x-2+5)=0#

or #(x-7)(x+3)=0#

i.e. #x=7# or #x=-3#