How do you solve #7=sqrt(12-x)+4# and identify any restrictions?
1 Answer
Apr 24, 2017
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"#
