can we presume what need to be proved? (1 Viewer)

Sirius Black

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There is one question in last year's PLC ext 1 trial -
"Suppose the roots of the equation X^3+pX^2+qX+r=0 are real. Show that the roots are in a geometric progression if q^3=p^2*r"

The solution was about to start with an assumption "Let the roots be (a/b), a, and ab. " and then they are trying to prove that q^3=p^2*r
However, from the question, wat we need 2 prove is the roots are (a/b), a, and ab by using q^3=p^2*r (that's my own interpretation =)
so is the solution incorrect ? Can we do it like the solution in HSC? :rolleyes:
 

withoutaface

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To prove something you can make the assumption of truth, and then see if it works when you substitute it in.
 

mojako

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For most polynomial questions, including this one, I think you can.

But if you can see a way to do it the better way, do it that way.
It is a good idea not to use the given result.
And BTW the word "show" doesn't necessarily mean you can use the given result. Sometimes they use this to mean "prove" - but you can sort of see that they mean prove when you read the question.
 

Archman

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there is also the "suffice to prove".
 

mojako

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Archman said:
there is also the "suffice to prove".
hmm.. like the natural growth thing where we differentiate stuff?
oh, in general, if the derivative of something is f(x), does it always mean that the primitive of f(x) is that something (ignoring possible constants)?
 

ngai

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Sirius Black said:
Can we do it like the solution in HSC? :rolleyes:
yes u can do it like that in the HSC
but i cant guarantee u dont get 0
 

withoutaface

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and instead of substituting and having them know you can generally just substitute down the bottom of the page and work your way up.
 

Slidey

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Sirius Black said:
so is the solution incorrect ? Can we do it like the solution in HSC? :rolleyes:
Only if you prove it by inspection.
 

Xayma

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withoutaface said:
and instead of substituting and having them know you can generally just substitute down the bottom of the page and work your way up.
That is the better method.

Because if you go let the roots be a/b, a, ab and then showing it, you aren't disproving that q<sup>3</sup>=p<sup>2</sup>r might also apply for an AP with a common difference of 2 etc.

You could show q<sup>3</sup>=p<sup>2</sup>r IFF the roots are in GP however.
 

Archman

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when i say "suffice to prove" it is strictly that.
eg: RTP a = b
u cant just work forward and get
|a| = |b| because that is not suffice to prove the first thing.
 

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