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

[Bushy]

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.

 

[MarkFL]

One method would be to equate:

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

Expand, and equate coefficients to get 3 equations in 3 unknowns.

 

[Bushy]

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?

 

[I like Serena]

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. ;)

 

[MarkFL]

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:

\(\displaystyle 8-a=2p+q\)

\(\displaystyle 16-8a=p^2+2pq\)

\(\displaystyle 5-16a=p^2q\)

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.