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- Thread starter Bushy
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- Mar 5, 2012

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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.

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The system I get is: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?

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

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

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

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.