Solving for 'a' in f(x) = 5: Limiting Solutions for f(x) = (x+4)^2(a-x)

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Discussion Overview

The discussion revolves around determining the value of 'a' in the function f(x) = (x+4)^2(a-x) such that the equation f(x) = 5 has exactly two solutions. The scope includes mathematical reasoning and problem-solving techniques related to polynomial equations.

Discussion Character

  • Mathematical reasoning, Technical explanation, Homework-related

Main Points Raised

  • One participant notes that f(x) has two solutions at x = a and x = -4, questioning the implications of moving the function down by 5 units.
  • Another participant suggests equating the function to a polynomial form and expanding it to derive a system of equations involving coefficients.
  • Several participants express concern over the complexity of the derived equations, indicating that the problem may be overly complicated for its perceived value in an exam context.
  • One participant proposes taking the derivative of f to find extrema, indicating that this could lead to a quicker solution by identifying where f(x) = 5 at another critical point.
  • Another participant reiterates the system of equations derived from the polynomial expansion, emphasizing the algebraic effort required to solve it and suggesting the remainder theorem as a potentially simpler approach.

Areas of Agreement / Disagreement

Participants express a mix of approaches to solving the problem, with no consensus on the best method. Some agree on the need for a derivative approach, while others focus on the algebraic system of equations, indicating a lack of agreement on the most efficient solution strategy.

Contextual Notes

The discussion reveals limitations in the clarity of the problem's requirements and the assumptions made about the function's behavior, particularly regarding the nature of its roots and extrema.

Bushy
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Consider the function

f(x) = (x+4)^2(a-x)

Determine 'a' such that f(x) = 5 has only two solutions.


I know f(x) has two solutions, x =a, -4.

Making f(x) = 5 moves the function down 5 units? then what?Regards.
 
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One method would be to equate:

$$(x+4)^2(a-x)-5=-(x-p)^2(x-q)$$

Expand, and equate coefficients to get 3 equations in 3 unknowns.
 
I have my 3 eqns with 3 unknowns, a-8 = 2p+q
8a-16 = =p^2-2pq
16a-5 = p^2q

but this seems quite a long way given it was only worth 2 marks on a previous exam.Is there an easier way to the end?
 
Welcome to MHB, bushy! :)

You're looking at a 3rd order polynomial that goes generally down.
It has a duplicate root at x=-4, meaning it has an extreme there.

To have exactly 2 solutions for f(x)=5, it needs to have another extreme at some x where f(x)=5.
Perhaps you can take the derivative of f and equate it to 0?
You should find x=-4 as a solution, and some other x that depends on a.
For that value of x, you should have f(x)=5.

It will bring you to the solution quicker. ;)
 
Bushy said:
I have my 3 eqns with 3 unknowns, a-8 = 2p+q
8a-16 = =p^2-2pq
16a-5 = p^2q

but this seems quite a long way given it was only worth 2 marks on a previous exam.Is there an easier way to the end?

The system I get is:

$$8-a=2p+q$$

$$16-8a=p^2+2pq$$

$$5-16a=p^2q$$

And you are correct that there is a bit of algebra involved in solving this system. To be honest, when I first saw this problem, I felt that applying the remainder theorem would most likely be the way to go.
 

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