How do you solve sqrt[x+2]=sqrt[4-x] and find any extraneous solutions?
1 Answer
Oct 10, 2017
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.
