Are you allowed to you calculus? You posted under algebra. Is there any restriction to what method is what I'm asking.I honestly don't know how to solve this. If it was a quadratic equation sure, but not this. View attachment 32343
No I know calculus. I just thought this was algebra. It wasn't stated.Are you allowed to you calculus? You posted under algebra. Is there any restriction to what method is what I'm asking.
Let's proceed with calculus then.No I know calculus. I just thought this was algebra. It wasn't stated.
can you explain this??2) f(x1)×f(x2)<0f(x_1)\times f(x_2)<0f(x1)×f(x2)<0.
f(x1)*f(x2)<0 is just another way of saying that f(x1) and f(x2) have different signs (one is negative and the other is positive)c
can you explain this??
works in what sense? when a = 0, x is either 0 or -3+sqrt(105)/2 or -3-sqrt(105)/2You need to find just one a such that the equation has three solutions.
I would first try to see if a=0 works
Calculus works just fine and you should(!!) know how to use calculus to solve this problem.
However, using trial and error can be faster and has less room for error.
Are you allowed to graph the function? If yes, then graph it without a in it and see if you need to raise or lower the graph to get three roots. Adjust a accordingly.
interesting, but why can't we have something like this (enjoy my paint drawing)View attachment 32344
Consider a generic cubic with 3 distinct roots. Notice that there are inflection points (condition 1). But for the function to cross the x-axis 3 times. One of them must be positive, and the other must be negative. In other words [imath]f(x_1)\times f(x_2)<0[/imath]
How many roots do you have in your picture?interesting, but why can't we have something like this (enjoy my paint drawing)View attachment 32345
I re-read the question and it say the function has three distinct roots if and only if a equals___?works in what sense? when a = 0, x is either 0 or -3+sqrt(105)/2 or -3-sqrt(105)/2
Now you have 3 turning points. It's no longer a cubic.
what about nowView attachment 32346
Steve, the solution is an interval, not a single value.I re-read the question and it say the function has three distinct roots if and only if a equals___?
This is saying that there is exactly one such a that works. This is not true. I bet that a=1/2 would work.
What you are drawing are not cubic equations. They are either 4th degree or 6 degree or 8th degree.... equations! You are clearly give a 3rd degree equation!!!!!!
what about nowView attachment 32346
I was wondering about that.Now you have 3 turning points. It's no longer a cubic.
Steve, the solution is an interval, not a single value.
from my experience, the question is formed in that way regardless of how many solutions there is. The goal is usually to see who will recognize that there are more solutions.I re-read the question and it say the function has three distinct roots if and only if a equals___?
This is saying that there is exactly one such a that works. This is not true. I bet that a=1/2 would work.
i suck at graphing equations...What you are drawing are not cubic equations. They are either 4th degree or 6 degree or 8th degree.... equations! You are clearly give a 3rd degree equation!!!!!!
okay so what i seem to not be understanding is, how do turning points dictate how many solutions for x we have?Now you have 3 turning points. It's no longer a cubic.
Steve, the solution is an interval, not a single value.