How do you solve (x-2)^2-25=0(x2)225=0?

3 Answers
Aug 24, 2016

x=7x=7

x=-3x=3

Explanation:

(x-2)^2-25=0(x2)225=0

or

(x-2)^2=25(x2)2=25

or

x-2=sqrt25x2=25

or

x-2=+-5x2=±5

or

x=2+-5x=2±5

or

x=2+5x=2+5

or

x=7x=7---------------------------------Ans 11

or

x=2-5x=25

or

x=-3x=3-----------------------------Ans 22

Aug 24, 2016

x=7,x=-3x=7,x=3

Explanation:

First step is to isolate the (x-2)^2(x2)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

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