Equation problem

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Loki123

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I honestly don't know how to solve this. If it was a quadratic equation sure, but not this. IMG_20220423_172034.jpg
 
Loki123 said:
No I know calculus. I just thought this was algebra. It wasn't stated.
Click to expand...
Let's proceed with calculus then.
For a cubic to have 3 distinct roots, it must satisfy these conditions:
1) Two real inflection points, [imath]x_1,x_2[/imath]
2) [imath]f(x_1)\times f(x_2)<0[/imath].

Find the inflection points.
What values of a will sastify this condition? [imath]f(x_1)\times f(x_2)<0[/imath], then you'll have your answer.
 
You 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.
 
Loki123 said:
c

can you explain this??
Click to expand...
Screen Shot 2022-04-23 at 11.44.34 AM.png
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]
 
Steven G said:
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.
Click to expand...

BigBeachBanana said:
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]
Click to expand...
interesting, but why can't we have something like this (enjoy my paint drawing)mati.png
 
Loki123 said:
works in what sense? when a = 0, x is either 0 or -3+sqrt(105)/2 or -3-sqrt(105)/2
Click to expand...
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.
 
Loki123 said:
:)
what about nowView attachment 32346
Click to expand...
Now you have 3 turning points. It's no longer a cubic.
Steven G said:
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.
Click to expand...
Steve, the solution is an interval, not a single value.
 
Steven G said:
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.
Click to expand...
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.
 
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