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Evaluate a+b+c

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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
Hi MHB,

I have solved the problem as stated below but I don't know if it's an unique solution and even if it is, I have no idea how to prove that would be the case.

Can anyone show me how to approach the problem correctly?

For the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$, all of its roots are positive real numbers. Evaluate the sum of $a+b+c$.

Attempt:

It's quite obvious from the values of the product of all 5 roots and the sum of them reveal that the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$ has roots of 2 and 4, of which 2 is the repeated root of multiplicity 4, since

$2+2+2+2+4=12$ and $2^4(4)=64$,

Then there are many ways to find the values for $a, b, c$ and at last, after the values of $a, b, c$ are known, we can conclude that $a+b+c=56+144-128=72$.
 

mente oscura

Well-known member
Nov 29, 2013
172
Hi MHB,


For the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$, all of its roots are positive real numbers. Evaluate the sum of $a+b+c$.
Hello.

[tex]Let \ r_1, \ r_2, \ r_3, \ r_4, \ r_5 \ roots \ of: \ /[/tex]

[tex]/ \ x^5-12x^4+ax^3+bx^2+c^x-64=(x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)[/tex]


[tex]a=r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5=C^5_2[/tex]

[tex]b=-(r_1r_2r_3+r_1r_2r_4+r_1r_2r_5+r_1r_3r_4+r_1r_3r_5+r_1r_4r_5+r_2r_3r_4+r_2r_3r_5+r_2r_4r_5+r_3r_4r_5)=C^5_3[/tex]

[tex]c=r_1r_2r_3r_4+r_1r_2r_3r_5+r_1r_2r_4r_5+r_1r_3r_4r_5+r_2r_3r_4r_5=C^5_4[/tex]

To include all combinations in the factors [tex]r_i[/tex], guarantees us that the solution is unique. Since you get all numerical products between 2, 3 and 4 factors, with the same result.

I do not know if I understood your question correctly.(Headbang)

regards.
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
Hello.

[tex]Let \ r_1, \ r_2, \ r_3, \ r_4, \ r_5 \ roots \ of: \ /[/tex]

[tex]/ \ x^5-12x^4+ax^3+bx^2+c^x-64=(x-r_1)(x-r_2)(x-r_3)(x-r_4)(x-r_5)[/tex]


[tex]a=r_1r_2+r_1r_3+r_1r_4+r_1r_5+r_2r_3+r_2r_4+r_2r_5+r_3r_4+r_3r_5+r_4r_5=C^5_2[/tex]

[tex]b=-(r_1r_2r_3+r_1r_2r_4+r_1r_2r_5+r_1r_3r_4+r_1r_3r_5+r_1r_4r_5+r_2r_3r_4+r_2r_3r_5+r_2r_4r_5+r_3r_4r_5)=C^5_3[/tex]

[tex]c=r_1r_2r_3r_4+r_1r_2r_3r_5+r_1r_2r_4r_5+r_1r_3r_4r_5+r_2r_3r_4r_5=C^5_4[/tex]

To include all combinations in the factors [tex]r_i[/tex], guarantees us that the solution is unique. Since you get all numerical products between 2, 3 and 4 factors, with the same result.
Thanks, mente oscura for the reply.:)

But...I don't quite get you especially the part when you mentioned the way to guarantee the only set values for all the 5 real positive roots (that I obtained via eyeballing) is the unique set of solution for solving the equation $x^5-12x^4ac^3+bx^2+cx-640=0$.

Yes, I know $a$ consists of the sum of $5\choose2$ terms, $b$ consists of the sum of $5\choose3$ terms and last, $c$ consists of the sum of $5\choose4$ terms, but how does one relate it to the number of sets of solution that we could get based on the only known values for the sum/product of roots?
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
Hi MHB,

I have solved the problem as stated below but I don't know if it's an unique solution and even if it is, I have no idea how to prove that would be the case.

Can anyone show me how to approach the problem correctly?

For the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$, all of its roots are positive real numbers. Evaluate the sum of $a+b+c$.

Attempt:

It's quite obvious from the values of the product of all 5 roots and the sum of them reveal that the equation $x^5-12x^4+ax^3+bx^2+cx-64=0$ has roots of 2 and 4, of which 2 is the repeated root of multiplicity 4, since

$2+2+2+2+4=12$ and $2^4(4)=64$,

Then there are many ways to find the values for $a, b, c$ and at last, after the values of $a, b, c$ are known, we can conclude that $a+b+c=56+144-128=72$.
That is not the only solution, and the value of $a+b+c$ is not unique. The equation $\displaystyle x^5 - 12x^4 + \frac{509}9x^3 - \frac{1174}9x^2 + \frac{440}3x - 64 = 0$ has roots $\displaystyle \frac43,\,2,\,\frac83,\,3,\,3$, and the sum of coefficients $a+b+c$ is $\displaystyle \frac{509}9 + \frac{440}3 - \frac{1174}9 = \frac{655}9 \ne 72.$
 
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anemone

MHB POTW Director
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Feb 14, 2012
3,756
That is not the only solution, and the value of $a+b+c$ is not unique. The equation $\displaystyle x^5 - 12x^4 + \frac{509}9x^3 - \frac{1174}9x^2 + \frac{440}3x - 64 = 0$ has roots $\displaystyle \frac43,\,2,\,\frac83,\,3,\,3$, and the sum of coefficients $a+b+c$ is $\displaystyle \frac{509}9 + \frac{440}3 - \frac{1174}9 = \frac{655}9 \ne 72.$
Thank you so much Opalg for your reply and also showing me the counter example (I have been trying very hard to find for another solution set by the help of wolfram, after I tried the combinations such as $\displaystyle \frac12,\,4,\,\frac52,\,r_4,\,r_5$ for which the remaining effort to look for the perfect candidates for all those 5 roots seemed no easy task for me.:mad:). Now I would just discard this question.:)
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I tried the combinations such as $\displaystyle \frac12,\,4,\,\frac52,\,r_4,\,r_5$
You are told that $r_1 + r_2 + r_3 + r_4 + r_5 = 12$ and $r_1r_2r_3r_4r_5 = 64$. The GM-AM inequality applied to the numbers $r_1,r_2,r_3,r_4,\sqrt{r_5},\sqrt{r_5}$ shows that $2 \leqslant \frac16(r_1 + r_2 + r_3 + r_4 + 2\sqrt{r_5})$, from which $r_5 \leqslant 2\sqrt{r_5}$, with equality only if $r_1=r_2=r_3=r_4=\sqrt{r_5}$. So if one of the roots is $4$ then the others must all be $2$. When I realised that, I tried putting three of the roots equal to $2,\,3$ and $3$, and I was surprised to find that I could then get a solution for the other two roots.
 
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anemone

MHB POTW Director
Staff member
Feb 14, 2012
3,756
You are told that $r_1 + r_2 + r_3 + r_4 + r_5 = 12$ and $r_1r_2r_3r_4r_5 = 64$. The GM-AM inequality applied to the numbers $r_1,r_2,r_3,r_4,\sqrt{r_5},\sqrt{r_5}$ shows that $2 \leqslant \frac16(r_1 + r_2 + r_3 + r_4 + 2\sqrt{r_5})$, from which $r_5 \leqslant 2\sqrt{r_5}$, with equality only if $r_1=r_2=r_3=r_4=\sqrt{r_5}$. So if one of the roots is $4$ then the others must all be $2$. When I realised that, I tried putting three of the roots equal to $2,\,3$ and $3$, and I was surprised to find that I could then get a solution for the other two roots.
Thank you again Opalg for your patience and willingness to teach me more about how to look for other possible solution for this problem. I really appreciate your help!:)

I like it how you made the six terms up $r_1,r_2,r_3,r_4,\sqrt{r_5},\sqrt{r_5}$ and then applied the AM-GM inequality for those numbers! I learn a great deal from you today! :eek: