Question

Can someone tell me if my work is an extraneous solution?The equation i needed to use wasa √x+b+c=dI already filled in the variables with the #’s i chose and I did the math but not sure if it’s an extraneous solution or not?

Answer 1

The given equation is

`a√(x+b+c)=d`

Where the values of the variables are given as,

`a=-3,b=6,c=8,d=16`

Substitute the given values into the above equation.

`-3√(x+6+8)=16`

Simplify the equation

`\begin{gathered} -3√(x+6+8)=16 \\ √(x+6+8)=(16)/(-3) \\ (√(x+6+8))^2=((16)/(-3))^2 \\ x+6+8=(256)/(9) \\ x+14=(256)/(9) \\ x=-14+(256)/(9) \\ x=(130)/(9)=14.44 \end{gathered}`

So, we got x = 14.44

Now let us talk about extraneous solutions.

Extraneous Solution:

An extraneous solution is a solution that we get by solving an equation but it is not really the solution meaning that it does not satisfy the equation!

For example:

For the given problem, we will plug the solution x = 14.44 into the original equation and check whether it satisfies the equation or not.

If it does not satisfy the equation then we have an extraneous solution

`\begin{gathered} -3√(x+6+8)=16 \\ -3√(14.44+6+8)=16 \\ -3√(28.44)=16 \\ -3(5.33291665039)=16 \\ -16=16 \end{gathered}`

As you can see, the left-hand side is not equal to the right-hand side of the equation.

This means that x = 14.44 does not satisfy the original equation so it is indeed an extraneous solution.

Note:

You have made an error (step 2) in the given picture. You cannot subtract something inside the square root, so first, you have to solve the square root (just like you did in step 6) and after that do such addition or subtraction operations.