Is this allowed? (1 Viewer)

[barbernator]

Ok this is from Sydney Grammar 2009.


Now my solution was different to in the solutions.

I differentiated once and stated that for a double root, f(a)=f'(a)=0

when solving for f'(x)=0 I obtained 2 results, x=+- root(p/3)

now to justify this I just said that one of them will be the double root, and regardless when substitution in they both yield the required result.

i know that isnt the best way to answer the question because it is ambiguous in the fact that you are proving the result with 1 correct double root and 1 that isnt a double root, yet they both yield the same result so can u just justify it by saying 1 must be the double root?

 

[asianese]

I had the same trouble too. Their solution was kind of left field compared to how you would normally do a question like that.

I think it is acceptable, as long as you justify it. They've also squared things and then square rooted, only taking the positive value.

 

[Trebla]

You can avoid this problem in the first place by a more indirect approach i.e. without actually specifying the double root itself
P(x) = x³ - px - q
P'(x) = 3x² - p
Suppose that x = a is a double root. A necessary condition for the double root is that P'(a) = P(a) = 0, hence it must be that
a² = p/3
and
a³ - ap - q = 0
a(a² - p) = q; but a² = p/3
=> -2ap/3 = q
=> 4a²p² = 9q²; but a² = p/3
Hence
27q² = 4p³