Root determined progression
x 3 − 9 x 2 + 2 3 x − 1 5 = 0
The roots of the equation above are all integers. What kind of sequence of numbers does these roots form?
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5 solutions
Would this mean that for a cubic function A x 3 + B x 2 + C x + D = 0 , − B = r 1 + r 2 + r 3 and − D = r 1 ∗ r 2 ∗ r 3 ?
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Yes. You are very near.. − B / A = r 1 + r 2 + r 3 a n d − D / A = r 1 ∗ r 2 ∗ r 3 a s p e r V i e t ′ s F o r m u l a .
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Thanks! Didn't know Vieta's formula applied for polynomials with degrees higher than two without searching it.
Can you please explain me what " number of changes of sign" has to do with nature of roots??
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The number of positive roots equals number of sign changes. So we know here that all roots are positive.
The roots must be divisors of 15. Hence the roots are from the integers 1,-1,3,-3,5,-5,15,-15. we find that x = 1 is a root of the equation and thus 0 = ( x − 1 ) ( x 2 − 8 x + 1 5 ) = ( x − 1 ) ( x − 3 ) ( x − 5 ) The roots are 1,3,5 Hence in A.P.
Moderator note:
Yes, the old fashion RRT works fine. But since it has hinted that it forms a certain kind of progression, it would be better to use Niranjan Khanderia's approach. It's would be impractical to use your method if the constant term is much larger.
You may also realise the fact that the sum of the coefficients is zero, and hence, ( x − 1 ) is a root of the given equation. Finding the rest of the roots from there on is straightforward.
you just have to realise that the sum of the coefficients is zero, and hence x = 1 is a root of this function. Then you can use the ruffini's rule to divide the polynomial by (x-1) and then we have: x^3 - 9x^2 +23x - 15 = (x-1)*(x^2 -8x +15) = 0 so solving the quadratic equation we'll have x = 3 or x = 5 so the roots are 1, 3 and 5 and they are in A.P.
The product of the three roots = 15
15 can be expressed only as the product of three integers , 1 , 3 and 5
So , the roots are
1 , 3 , 5 (Arithmetic progression )
The sum of roots = 9 = 1 + 3 + 5 .....................................(just to make sure0
Just curious. What is an arithmetic geometric Progression?
It is a series in which the numerator of each term is in arithmetic progression and the denominator of each term is in geometric progression.
For example
1 + 3/4 + 5/16 + 7/64 + 9/256 + 11/1024 +..................
Here the numerators are in A.P. whereas the denominators are in G.P.
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How do we deal with such a arithmetic geometric Progression?
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Let me solve the example that I have mentioned.
Let S = 1 + 3/4 + 5/16 + 7/64 + 9/256 + 11/1024 +.................. ----- (i)
S/4 = 1/4 + 3/16 + 5/64 + 7/256 + 9/1024 + ......................... ------(ii)
( Here multiply S by geometric ratio r =1/4 )
Subtracting, eq (ii) from eq (i) , we get,
3S/4 = 1+2/4 + 2/16 + 2/64 + 2/256 +2/1024 ..................
=> 3S/4 = 1 + [ ( 2/4 ) / ( 1-1/4) ] => 3S/4 = 1 + 2/3 = 5/3 => S = 5/3 * 4/3 = 20/9
Hope it helps @Niranjan Khanderia Sir
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@Manish Dash – Thank you a lot for the help and promptness.
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@Niranjan Khanderia – That is my pleasure. By the way are you really 87 years old ?
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@Manish Dash – Yes. Why you doubted? 1928 January.
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@Niranjan Khanderia – I had doubt because there are some people who do not show their real age on the internet and also in Brilliant. I really admire and congratulate you for your intelligence as there are very few people in this world who are fit enough to do mathematics and access internet at your age. I apologise for my query. You are like my grandparent, @Niranjan Khanderia Sir
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@Manish Dash – Glad to. Recive your reply and sentiments. I love maths from young age. I feel very happy seeing you teenagers knowing much more that what I do. Wish you the best.
Roots are integers. Three changes of signs. So all roots +tive.. -15 means r 1 ∗ r 2 ∗ r 3 = 1 5 = 1 ∗ 3 ∗ 5 . A l s o r 1 + r 2 + r 3 = 9 . Clearly roots are 1,3,5 and they are in A.P.