How do you solve #1/2(x-4)^2+3=11#?

1 Answer
Jim G.
May 7, 2017

#x=0" or " x=8#

Explanation:

#color(blue)"Isolate " (x-4)^2" on the left side"#

#"subtract 3 from both sides"#

#1/2(x-4)^2cancel(+3)cancel(-3)=11-3#

#rArr1/2(x-4)^2=8#

#"multiply both sides by 2"#

#cancel(2)xx1/cancel(2)(x-4)^2=2xx8#

#rArr(x-4)^2=16larr(x-4)^2" isolated on left"#

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

#sqrt((x-4)^2)=color(red)(+-)sqrt16larr" note plus or minus"#

#rArrx-4=+-4#

#"add 4 to both sides"#

#xcancel(-4)cancel(+4)=+-4+4#

#rArrx=4+-4larrcolor(red)" 2 solutions"#

#rArrx_1=4-4=0" or " x_2=4+4=8#

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