we are given the following expression:
[tex]a\sqrt[]{x+b}+c=d[/tex]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:
[tex]\begin{gathered} d=4 \\ c=8 \\ a=2 \\ b=-5 \end{gathered}[/tex]We get the equation:
[tex]2\sqrt[]{x-5}+8=4[/tex]Now let's take the following values for the constants:
[tex]\begin{gathered} c=4 \\ d=8 \\ a=2 \\ b=-5 \end{gathered}[/tex]We get the equation:
[tex]2\sqrt[]{x-5}+4=8[/tex]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:
[tex]2\sqrt[]{x-5}=4-8[/tex]Now we'll divide by "2":
[tex]\sqrt[]{x-5}=\frac{4-8}{2}[/tex]Let's choose the following values:
[tex]\begin{gathered} d=4 \\ c=8 \\ a=2 \\ b=-5 \end{gathered}[/tex]Now let's solve for "x":
[tex]\sqrt[]{x-5}=-\frac{4}{2}[/tex][tex]\sqrt[]{x-5}=-2[/tex]Elevating both sides to the second power:
[tex](\sqrt[]{x-5})^2=(-2)^2[/tex]Solving:
[tex]x-5=4[/tex]Adding 5 on both sides:
[tex]\begin{gathered} x=4+5 \\ x=9 \end{gathered}[/tex]Now that we get a solution we need to check it by replacing the value we found for "x" in the initial equation:
[tex]\sqrt[]{x-5}=-2[/tex]Replacing the value of "x":
[tex]\sqrt[]{9-5}=-2[/tex]Solving the operation inside the radical:
[tex]\sqrt[]{4}=-2[/tex]Solving the radical:
[tex]2=-2[/tex]Now we use the second equation:
[tex]2\sqrt[]{x-5}+4=8[/tex]Isolating the radical we get
[tex]\sqrt[]{x-5}=\frac{8-4}{2}[/tex]Solving the operations:
[tex]\begin{gathered} \sqrt[]{x-5}=\frac{4}{2} \\ \sqrt[]{x-5}=2 \end{gathered}[/tex]Squaring both sides:
[tex](\sqrt[]{x-5})^2=(2)^2[/tex]Solving the square:
[tex]x-5=4[/tex]adding 5 on both sides:
[tex]\begin{gathered} x=4+5 \\ x=9 \end{gathered}[/tex]Now, replacing the value of "x" in the original equation:
[tex]\sqrt[]{9-5}=2[/tex]Solving the operation inside the radical:
[tex]\sqrt[]{4}=2[/tex]Solving the radical.
[tex]2=2[/tex]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.
