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.

Guest Feb 28, 2018

#1**+1 **

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.

ElectricPavlov
Feb 28, 2018

#1**+1 **

Best Answer

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.

ElectricPavlov
Feb 28, 2018