Equation problem

[Steven G]

Since you want a range, then BBB is 100% correct with his method!
Now look carefully at the graph in post #10. Assuming a = 0 in this graph how much can we raise and lower this graph so it will still have 3 roots? That is what this entire problem is asking.

 

[Steven G]

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.

I think you meant relative max/min points and not inflection points

 

[Steven G]

Based on post #10 and that the graph is using a=0, then what would the answer be?

 

[Loki123]

Based on post #10 and that the graph is using a=0, then what would the answer be?

x1,x2,0

 

[BigBeachBanana]

I think you meant relative max/min points and not inflection points

Yes, I misspoke. These should be critical points.

okay so what i seem to not be understanding is, how do turning points dictate how many solutions for x we have?

I showed you in post#10 a generic cubic with 3 distinct roots. You see that it must have 2 critical points. Further more f(x1)>0 and f(x2)<0 so the function f(x) itself crosses the x-axis 3 times. Hence, giving you 3 distinct roots.

 

[Steven G]

x1,x2,0

That is not a valid answer. I thought that we agreed that we are looking for a range for a. The form of the answer would be t<a<r.

Exactly what did you mean by x1,x2,0?? Just curious

 

[Loki123]

That is not a valid answer. I thought that we agreed that we are looking for a range for a. The form of the answer would be t<a<r.

Exactly what did you mean by x1,x2,0?? Just curious

those are the cordinates of x when y=0 and I thought you that's what you were asking for.
i guess x1<a<x2

 

[Steven G]

those are the cordinates of x when y=0 and I thought you that's what you were asking for.
i guess x1<a<x2

No guessing! Please try again with thinking. BTW, I did not say that your answer is right or wrong. I just don't accept answer that came from a guess.

 

[Loki123]

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.

i think it's x1<a<x2

 

[Steven G]

those are the cordinates of x when y=0 and I thought you that's what you were asking for.
i guess x1<a<x2

Wait a minute. You think that when x=x1, the y value of the graph is 0? Same for x=x2.

 

[Steven G]

i think it's x1<a<x2

OK, Now go and find x1 and x2 in your function. Remember, we are using a=0 to find x1 and x2.

 

[Steven G]

BTW, which is larger, x1 or x2?

 

[Loki123]

Wait a minute. You think that when x=x1, the y value of the graph is 0? Same for x=x2.

well now that you emphasized it like that, i looked better at it and i realized it's not correct because the function doesn't cross y right under x1 and x2

 

[Loki123]

BTW, which is larger, x1 or x2?

i feel like this is a trick question, but i think x2.

 

[Loki123]

OK, Now go and find x1 and x2 in your function. Remember, we are using a=0 to find x1 and x2.

1 and -2, now what?

 

[BigBeachBanana]

1 and -2, now what?

Second condition

 

[Loki123]

Second condition

f(1)=-7
f(-2)=20
so a(-7,20)??

 

[BigBeachBanana]

f(1)=-7
f(-2)=20
so a(-7,20)??

Yes [imath]a \in (-7,20)[/imath].

 

[Loki123]

Yes [imath]a \in (-7,20)[/imath].

When would it be [-7,20] ?

 

[BigBeachBanana]

When would it be [-7,20] ?

Plug in a=-7 and a=20. How many roots do you get?
https://www.desmos.com/calculator
Drag or play the slider far different value of a.