How do you solve # sqrt[x+2]=sqrt[4-x]# and find any extraneous solutions?

1 Answer
Oct 10, 2017

Answer:

#x=1#

Explanation:

#"to access the contents of the radicals"#

#color(blue)"square both sides"#

#•color(white)(x)sqrtaxxsqrta=(sqrta)^2=a#

#(sqrt(x+2))^2=(sqrt(4-x))^2#

#rArrx+2=4-x#

#"add x to both sides"#

#x+x+2=4cancel(-x)cancel(+x)#

#rArr2x+2=4#

#"subtract 2 from both sides"#

#2xcancel(+2)cancel(-2)=4-2#

#rArr2x=2#

#"divide both sides by 2"#

#(cancel(2) x)/cancel(2)=2/2#

#rArrx=1#

#color(blue)"As a check"#

Substitute this value into the equation and if both sides are equal then it is the solution.

#"left side "=sqrt(1+2)=sqrt3#

#"right side "=sqrt(4-1)=sqrt3#

#rArrx=1" is the solution"#

There are no other solutions hence no extraneous solutions.