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

3 Answers
Mar 7, 2017

Answer:

#"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

Answer:

# 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.

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Mar 7, 2017

Answer:

#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"#