How do you solve #7=sqrt(12-x)+4# and identify any restrictions?

1 Answer
Apr 24, 2017

Answer:

#x=3#

Explanation:

#color(blue)"Isolate the root ""by subtracting 4 from both sides"#

#7-4=sqrt(12-x)cancel(+4)cancel(-4)#

#rArrsqrt(12-x)=3#

#"to 'undo' the root "color(blue)"square both sides"#

#(sqrt(12-x))^2=3^2#

#rArr12-x=9#

#"subtract 12 from both sides"#

#cancel(12)cancel(-12)-x=9-12#

#rArr-x=-3rArrx=3#

#color(blue)"As a check"#

substitute this value into the right side of the equation and if equal to the left side then it is the solution.

#"right side "=sqrt(12-3)+4=sqrt9+4=3+4=7#

#rArrx=3" is the solution"#