MHB Find the sum of all real numbers

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The discussion focuses on finding the sum of all real numbers \( a \) that satisfy the polynomial equation \( 5a^4-10a^3+10a^2-5a-11=0 \). Participants analyze the polynomial's structure, noting its Galois group and implications for the roots. A transformation of the polynomial suggests that the two real roots sum to 1. Corrections are made regarding the focus on real roots only, confirming that the final sum of the real solutions is indeed 1. The conversation highlights the collaborative nature of problem-solving in mathematics.
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Find the sum of all real numbers $a$ such that $5a^4-10a^3+10a^2-5a-11=0$.
 
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anemone said:
Find the sum of all real numbers $a$ such that $5a^4-10a^3+10a^2-5a-11=0$.

[sp]$$P(a)=5a^4-10a^3+10a^2-5a-11=5(a^2-a)^2+5(a^2-a)-11$$[/sp]

...and the rest is but "sound and fury signifying nothing"

.
 
Last edited:
zzephod said:
[sp]$$P(a)=5a^4-10a^3+10a^2-5a-11=5(q^2-1)^2+5(a^2-a)-11$$[/sp]

...and the rest is but "sound and fury signifying nothing"

.

Perhaps you mean:

[sp]$$P(a)=5a^4-10a^3+10a^2-5a-11=5\left(a(a-1)\right)^2+5a(a-1)-11$$[/sp]

Now you are set to finish...
 
Hey! I actually know how to do this one! Wheeee!
[math]5 a^4 - 10 a^3 + 10 a^2 - 5 a - 11 = 0[/math]

This is a quartic equation, so let's do what we all do when presented by such a polynomial equation: Get rid of the cubic term. (Well, okay, the second thing anyway. I suspect the first thing to do is go to Wolfram|Alpha to get the answer and work backward.)

Anyway, let a = b + 1/2.
[math]5 \left ( b + \frac{1}{2} \right )^4 - 10 \left ( b + \frac{1}{2} \right )^3 + 10 \left ( b + \frac{1}{2} \right )^2 - 5 \left ( b + \frac{1}{2} \right ) - 11 = 0[/math]

Expanding out and simplifying reveals we also get to get rid of the linear term:
[math]5b^4 - \frac{5}{2}b^2 - \frac{191}{16} = 0[/math]

Solving this I get
[math]b = \pm\sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} }[/math]

[math]a = \pm \sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} } + \frac{1}{2}[/math]

(Note: The two +/- are unrelated.)

Now, list all the possible values of a (not all real) and add them all up.

[math]\text{Sum of a} = \left ( \sqrt{ -\frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( i \cdot \sqrt{ \frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( - \sqrt{ -\frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( -i \cdot \sqrt{ \frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right )[/math]

[math]= 4 \cdot \frac{1}{2} = 2[/math]

=-Dan
 
topsquark said:
Hey! I actually know how to do this one! Wheeee!
[math]5 a^4 - 10 a^3 + 10 a^2 - 5 a - 11 = 0[/math]

This is a quartic equation, so let's do what we all do when presented by such a polynomial equation: Get rid of the cubic term. (Well, okay, the second thing anyway. I suspect the first thing to do is go to Wolfram|Alpha to get the answer and work backward.)

Anyway, let a = b + 1/2.
[math]5 \left ( b + \frac{1}{2} \right )^4 - 10 \left ( b + \frac{1}{2} \right )^3 + 10 \left ( b + \frac{1}{2} \right )^2 - 5 \left ( b + \frac{1}{2} \right ) - 11 = 0[/math]

Expanding out and simplifying reveals we also get to get rid of the linear term:
[math]5b^4 - \frac{5}{2}b^2 - \frac{191}{16} = 0[/math]

Solving this I get
[math]b = \pm\sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} }[/math]

[math]a = \pm \sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} } + \frac{1}{2}[/math]

(Note: The two +/- are unrelated.)

Now, list all the possible values of a (not all real) and add them all up.

[math]\text{Sum of a} = \left ( \sqrt{ -\frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( i \cdot \sqrt{ \frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( - \sqrt{ -\frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( -i \cdot \sqrt{ \frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right )[/math]

[math]= 4 \cdot \frac{1}{2} = 2[/math]

=-Dan

Not a solution but

as sum of 2 complex roots are 1 so the sum of 2 real roots are 1
 
I would like to point out that topsquark's "Tschrinahusen transform doing more than it should" is not a coincidence. The polynomial in the OP has Galois group $D_4$, which is of order $8$. As

$$|\mathbf{Gal}(K/F)| = [K : F]$$

We know that the field extension over $\Bbb Q$ constructed by adjoining the roots of the polynomial is of degree $8$. But if there were any cubic element in the extension, then by $[K : F] = [K : E][E : F]$ for some $E/F$ generated by the conjugates of the cubic (that'd be atmost of degree $6$) $8$ would be divisible by $3$, a contradiction. Thus there is no cubic radicals appearing in the expressions of the roots of the quartic, hence there is a transformation $p(x) \leadsto p'(x)$ where $p(x)$ is the quartic in OP, going to $p'(x)$, a quadratic in quartic disguise.

In particular the resolvent cubic (I have mentioned what they are in another post) factors in a linear term and a quadratic term.

Just sayin'.
 
Last edited:
[sp]The graph of the function shows two real roots (one positive and one negative) whose sum looks suspiciously close to $1$. So I wondered whether there would be a factorisation of the form $$5x^4-10x^3+10x^2-5x-11= 5(x^2 - x - \alpha)(x^2- x + \beta).$$ Comparing coefficients of powers of $x$, that will hold provided that $$5(\beta + 1 - \alpha) = 10 \qquad( \text{coefficient of }x^2),$$ $$5( \alpha - \beta) = -5 \qquad( \text{coefficient of }x),$$ $$5\alpha\beta = 11 \qquad( \text{constant term}).$$ The first two equations both reduce to $\alpha - \beta + 1 = 0$. Substituting the value of $\beta$ from the third equation gives $\alpha - \dfrac{11}{5\alpha} + 1 = 0,$ a quadratic equation with a positive root $\alpha = \dfrac{4\sqrt5 - 5}{10}$, the corresponding value of $\beta$ being $\beta = 2(4\sqrt5 + 5)$.

The "$\alpha$" factor then shows that the two real roots have sum $1$. (The "$\beta$" factor shows that the two nonreal roots also have sum $1$.)[/sp]
 
MarkFL said:
Perhaps you mean:

[sp]$$P(a)=5a^4-10a^3+10a^2-5a-11=5\left(a(a-1)\right)^2+5a(a-1)-11$$[/sp]

Now you are set to finish...

It now reads as I had intended, not quite sure how the transcription error could have occurred in the form that it did.

.
 
topsquark said:
Hey! I actually know how to do this one! Wheeee!
[math]5 a^4 - 10 a^3 + 10 a^2 - 5 a - 11 = 0[/math]

This is a quartic equation, so let's do what we all do when presented by such a polynomial equation: Get rid of the cubic term. (Well, okay, the second thing anyway. I suspect the first thing to do is go to Wolfram|Alpha to get the answer and work backward.)

Anyway, let a = b + 1/2.
[math]5 \left ( b + \frac{1}{2} \right )^4 - 10 \left ( b + \frac{1}{2} \right )^3 + 10 \left ( b + \frac{1}{2} \right )^2 - 5 \left ( b + \frac{1}{2} \right ) - 11 = 0[/math]

Expanding out and simplifying reveals we also get to get rid of the linear term:
[math]5b^4 - \frac{5}{2}b^2 - \frac{191}{16} = 0[/math]

Solving this I get
[math]b = \pm\sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} }[/math]

[math]a = \pm \sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} } + \frac{1}{2}[/math]

(Note: The two +/- are unrelated.)

Now, list all the possible values of a (not all real) and add them all up.

[math]\text{Sum of a} = \left ( \sqrt{ -\frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( i \cdot \sqrt{ \frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( - \sqrt{ -\frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right ) + \left ( -i \cdot \sqrt{ \frac{1}{4} + \frac{7 \sqrt{5}}{10} } + \frac{1}{2} \right )[/math]

[math]= 4 \cdot \frac{1}{2} = 2[/math]

=-Dan

Edit: anemone has just let me know that I have a small error in my answer. The question was originally cast as "the sum of all real a" and I added in the sum of all a's. In other words the complex a's have to be discarded and not be put into the sum. So there are only two numbers a that we can add. The corrected sum is thus [math]2 \cdot 2 = 1[/math]

-Dan
 
  • #10
Thanks all for participating and I believe this thread somehow reassures us how one math problem usually can be tackled using several different approaches.
 
  • #11
zzephod said:
It now reads as I had intended, not quite sure how the transcription error could have occurred in the form that it did.

.

We are all human, and can make errors, particularly in haste. I know I've made my share. :D

We do ask in our http://mathhelpboards.com/challenge-questions-puzzles-28/guidelines-posting-answering-challenging-problem-puzzle-3875.html though that complete solutions be given to problems here in our "Challenge Questions and Puzzles" forum.
 

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