WholeLottaLove wrote:
Which of the following represents all the possible values of x that are solutions to the equation 3x=|x^2 -10| ?
x>3.33:
Positive:
3x=x^2-10
-x^2+3x+10=0
x^2-3x-10=0
(x-5)*(x+2)=0
x=5, x=-2
x=5 falls in the range of x>3.33. X=2 does not.
x<3.33
Negative:
3x=-(x^2-10)
3x=-x^2+10
x^2+3x-10=0
(x+5)*(x-2)=0
x=-5, x=2
Both -5 and 2 fall within the range of x<3.33
Solutions = -5,2,5
In other words, in a similar problem I would check the results of the positive and negative case and see if they fell within the range I was testing. For example, for x>3.33, we get two solutions (x=5, x=-2) but only x=5 is valid. Why is that? I see that you are using 3x (and therefore, x) as the metric for what values are valid and what ones arent) but it doesn't seem like thats the case in other problems)
Your intervals are not correct:
\(3x=|x^2 -10|\), \(x^2 -10>0\) if \(x>\sqrt{10}\) and if \(x<-\sqrt{10}\).
So if \(x>\sqrt{10}\) or \(x<-\sqrt{10}\) => positive
\(3x=x^2-10\) that has two solutions: \(x=5\) (possible) and \(x=-2\) not possible because it's outside the range we are considering.
If \(-\sqrt{10}<x<\sqrt{10}\) => negative
\(3x=-x^2+10\) that has two solutions: \(x=2\) (possible) and \(x=-5\) not possible because it's outside the range we are considering.
So overall x can be 2 or 5.
"Why is that? " you anlyze parts of the function each time, so you have to pay attention to which ranges you are considering. If a given solution falls within that range, then it's solution, but it it falls out, it's not valid.
Hope it clarifies