To determine which solution to the equation is extraneous, we first must clarify the equation. It appears there may have been a typographical error, and we interpret it as:
1x + 1x^2 + 2 = 2
Next, let’s simplify the equation:
1x + 1x^2 + 2 – 2 = 0
Which simplifies to:
x^2 + x = 0
This can be factored as:
x(x + 1) = 0
From this, we find the solutions:
- x = 0
- x = -1
Now, we should check both solutions in the original equation to see if either may be extraneous. An extraneous solution is one that arises from the solving process but does not satisfy the original equation.
For x = 0:
1(0) + 1(0^2) + 2 = 2
This simplifies to:
0 + 0 + 2 = 2
This is correct.
For x = -1:
1(-1) + 1(-1^2) + 2 = 2
This simplifies to:
-1 + 1 + 2 = 2
This is also correct.
Since both values satisfy the original equation, we conclude that neither solution is extraneous. However, if there were additional constraints or if the equation had been set up differently leading us to an impossible solution when checking back with the original equation, that’s where we would find an extraneous solution.