NVM! I solved this question by solving for x, thank you for helping, melody!
Here is the graph
Using the graph can you see what the answer is?
You also need to be able to work these out without graphing first (it is good to graph afterwards for checking you answer)
X^2 is always going to be positive.
x can't be 1
Do you know how to use calculus Tom?
If so it is easy just to differentiate and then set y'=0 to find the stationary points etc.
what caculus? uhhh NO im in 8th GRADE! Im supposed to be in 6th!
It doesn't actually matter what the solution to this quadratic is. The only thing that is important is that we find all values such that this quadratic (in ) has a real solution. For that we look at the discriminant.
We see that we can only find a suitable value of x if \( y^2 + 4y \ge 0\)
You need to use more explanation, I have no idea what you are talking about.