Find the solution of 4 times the square root of the quantity of x plus 2 equals negative 16, and determine if it is an extraneous solution.
+36915 x
x+2
sqrt(x+2)
4 * sqrt(x+2)
4* sqrt(x+2) = -16 There are no real solutions to this as we cannot, by convention, get a negative sqrt
square both sides 16 (x+2) = 256
(x+2) = 16
x=14 Extraneous
From MathIsFun website:
Example: you have an equation and after some work come up with two roots (where it equals zero) "a" and "b". When you put "a" into the original equation it becomes zero, but when you put in "b" it doesn't. So "b" is an extraneous root. This often happens when we square both sides during our solution.
+36915 x
x+2
sqrt(x+2)
4 * sqrt(x+2)
4* sqrt(x+2) = -16 There are no real solutions to this as we cannot, by convention, get a negative sqrt
square both sides 16 (x+2) = 256
(x+2) = 16
x=14 Extraneous
From MathIsFun website:
Example: you have an equation and after some work come up with two roots (where it equals zero) "a" and "b". When you put "a" into the original equation it becomes zero, but when you put in "b" it doesn't. So "b" is an extraneous root. This often happens when we square both sides during our solution.
