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Let \(k\) and \(m\) be real numbers, and suppose that the roots of the equation

 

\(x^3 - 7x^2 + kx - m = 0\)

 

are three distinct positive integers. Compute \(k+m\)

 Feb 21, 2020
 #1
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Let the roots be p, q and r, and show that p + q + r = 7.

If p, q and r are distinct positive integers, they have to be 1, 2 and 4.

 Feb 22, 2020
 #2
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Ohh, I see. I haven't grasped Vietas yet.

 Feb 22, 2020
 #3
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Now that I think about it, why do p+q+r=7?

 Feb 22, 2020
 #4
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If the roots are p, q and r then:

 

(x-p)(x-q)(x-r) = x^3 - (p+q+r)x^2 + (pq+pr+qr)x - pqr

 

Compare with the x^2 term of the original cubic to see that p+q+r = 7

.

Alan  Feb 23, 2020
 #5
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ohh, ok, so would we get k+m=7?

 Feb 24, 2020
 #6
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No! Look at Guest #1's reply.  Set p = 1, q = 2, r = 4, then work out k and m by comparing with the x term and the constant term.

Alan  Feb 25, 2020
edited by Alan  Feb 25, 2020
 #7
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ahh, Ok I see. Got it, k+m=22. I get how to do it.

 Feb 25, 2020

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