Question

Hello. I need help on the questions below. While this is not for a grade, I do need to get it right to move on in the class. I submitted this before and this is the feedback I got:"you need in each equation to have a constant for a, b, c, and d, not any zeros. The (x+b) terms have to be under the radical. the extraneous one has to be set up so that when you isolate the radical term on the left side, the right side is negative."I don't know if you can make anything of that, but any help is appreciated!!

Answer 1

we are given the following expression:

`a\sqrt[]{x+b}+c=d`

We are asked to find constants a, b, c, and d such that we get an extraneous solution and a non-extraneous solution.

Let's remember that an extraneous solution arises when solving a problem we reduce it to a simpler problem and get a solution but when replacing that solution in reality it's not a solution to the problem because it is undetermined or outside the domain of the original problem.

Part 1. Let's choose the following values:

`\begin{gathered} d=4 \\ c=8 \\ a=2 \\ b=-5 \end{gathered}`

We get the equation:

`2\sqrt[]{x-5}+8=4`

Now let's take the following values for the constants:

`\begin{gathered} c=4 \\ d=8 \\ a=2 \\ b=-5 \end{gathered}`

We get the equation:

`2\sqrt[]{x-5}+4=8`

Part 2. To get the extraneous solution we will isolate the radical first from the expression. To do that we will subtract "8" from both sides:

`2\sqrt[]{x-5}=4-8`

Now we'll divide by "2":

`\sqrt[]{x-5}=(4-8)/(2)`

Let's choose the following values:

`\begin{gathered} d=4 \\ c=8 \\ a=2 \\ b=-5 \end{gathered}`

Now let's solve for "x":

`\sqrt[]{x-5}=-(4)/(2)`
`\sqrt[]{x-5}=-2`

Elevating both sides to the second power:

`(\sqrt[]{x-5})^2=(-2)^2`

Solving:

`x-5=4`

Adding 5 on both sides:

`\begin{gathered} x=4+5 \\ x=9 \end{gathered}`

Now that we get a solution we need to check it by replacing the value we found for "x" in the initial equation:

`\sqrt[]{x-5}=-2`

Replacing the value of "x":

`\sqrt[]{9-5}=-2`

Solving the operation inside the radical:

`\sqrt[]{4}=-2`

Solving the radical:

`2=-2`

Now we use the second equation:

`2\sqrt[]{x-5}+4=8`

Isolating the radical we get

`\sqrt[]{x-5}=(8-4)/(2)`

Solving the operations:

`\begin{gathered} \sqrt[]{x-5}=(4)/(2) \\ \sqrt[]{x-5}=2 \end{gathered}`

Squaring both sides:

`(\sqrt[]{x-5})^2=(2)^2`

Solving the square:

`x-5=4`

adding 5 on both sides:

`\begin{gathered} x=4+5 \\ x=9 \end{gathered}`

Now, replacing the value of "x" in the original equation:

`\sqrt[]{9-5}=2`

Solving the operation inside the radical:

`\sqrt[]{4}=2`

Solving the radical.

`2=2`

Therefore, x = 9 is a solution to this equation.

Part 3. Since the value we found for "x" in the first equation does not give a solution, this means that x = 9 is an extraneous solution for the first given values of the constants.