Question

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

Answer 1

`"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"`

Answer 2

`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=4` `rarr` Straight line parallel to x-axis.

Answer 3

`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"`