How do you solve (x-5)^2=4?

3 Answers
Mar 7, 2017

"the answer : x=3 or x=7"

Explanation:

(x-5)^2=4

sqrt ((x-5)^2)=sqrt 4

+- (x-5)=+-2

"if "+(x-5)=+2" then "x-5=2" , "x=7

"if "+(x-5)=-2" then "x-5=-2" , "x=3

"if "-(x-5)=+2" then "-x+5=2" , "x=3

"if "-(x-5)=-2" then "-x+5=-2" , "x=7

"thus ;"

"x=3 or x=7"

Mar 7, 2017

x= 7 & x= 3

Explanation:

  • Method 1-

(x-5)^2= 4

rArr x-5= +-2 ----(taking square root both sides)

rArr x-5= 2 & rArr x-5= -2

rArr x= 7 & rArr x= 3

  • Method 2-

(x-5)^2= 4

rArrx^2-10x+25= 4

rArrx^2-10x+21= 0

rArrx^2-7x-3x+21= 0

rArrx(x-7)-3(x-7)=0

rArr(x-3)(x-7)=0

rArr x= 7 & rArr x= 3

  • Method 3-

Graphical Approach

Plot y=(x-5)^2 and y=4 on graph paper. Their intersection points are its solutions.

y=(x-5)^2 rarr Parabola
y=4rarr Straight line parallel to x-axis.

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Mar 7, 2017

x=3" or "x=7

Explanation:

To 'undo' the square on the left side.

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

sqrt((x-5)^2)=+-sqrt4

rArrx-5=+-2larrcolor(red)" there are 2 solutions"

• x-5=+2

rArrx=2+5

rArrx=7to(color(red)(1))

• x-5=-2

rArrx=-2+5

rArrx=3to(color(red)(2))

color(blue)"As a check"

Substitute these values into the left side of the equation and if equal to the right side then they are the solutions.

x=7to(7-5)^2=4=" right side"

x=3to(3-5)^2=4=" right side"

rArrx=3" or "x=7" are the solutions"