Find the sum of all real numbers

[anemone]

Find the sum of all real numbers $a$ such that $5a^4-10a^3+10a^2-5a-11=0$.

 

[zzephod]

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"

.

 

[MarkFL]

[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...

 

[topsquark]

Hey! I actually know how to do this one! Wheeee!

Spoiler

\(\displaystyle 5 a^4 - 10 a^3 + 10 a^2 - 5 a - 11 = 0\)

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.
\(\displaystyle 5 \left ( b + \frac{1}{2} \right )^4 - 10 \left ( b + \frac{1}{2} \right )^3 + 10 \left ( b + \frac{1}{2} \right )^2 - 5 \left ( b + \frac{1}{2} \right ) - 11 = 0\)

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

Solving this I get
\(\displaystyle b = \pm\sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} }\)

\(\displaystyle a = \pm \sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} } + \frac{1}{2}\)

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

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

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

\(\displaystyle = 4 \cdot \frac{1}{2} = 2\)

=-Dan

 

[kaliprasad]

Hey! I actually know how to do this one! Wheeee!

Spoiler

\(\displaystyle 5 a^4 - 10 a^3 + 10 a^2 - 5 a - 11 = 0\)

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.
\(\displaystyle 5 \left ( b + \frac{1}{2} \right )^4 - 10 \left ( b + \frac{1}{2} \right )^3 + 10 \left ( b + \frac{1}{2} \right )^2 - 5 \left ( b + \frac{1}{2} \right ) - 11 = 0\)

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

Solving this I get
\(\displaystyle b = \pm\sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} }\)

\(\displaystyle a = \pm \sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} } + \frac{1}{2}\)

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

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

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

\(\displaystyle = 4 \cdot \frac{1}{2} = 2\)

=-Dan

Not a solution but

Spoiler

as sum of 2 complex roots are 1 so the sum of 2 real roots are 1

 

[mathbalarka]

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'.

 

[Opalg]

[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]

 

[zzephod]

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]

Hey! I actually know how to do this one! Wheeee!

Spoiler

\(\displaystyle 5 a^4 - 10 a^3 + 10 a^2 - 5 a - 11 = 0\)

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.
\(\displaystyle 5 \left ( b + \frac{1}{2} \right )^4 - 10 \left ( b + \frac{1}{2} \right )^3 + 10 \left ( b + \frac{1}{2} \right )^2 - 5 \left ( b + \frac{1}{2} \right ) - 11 = 0\)

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

Solving this I get
\(\displaystyle b = \pm\sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} }\)

\(\displaystyle a = \pm \sqrt{ -\frac{1}{4} \pm \frac{7 \sqrt{5}}{10} } + \frac{1}{2}\)

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

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

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

\(\displaystyle = 4 \cdot \frac{1}{2} = 2\)

=-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 \(\displaystyle 2 \cdot 2 = 1\)

-Dan

 

[anemone]

Thanks all for participating and I believe this thread somehow reassures us how one math problem usually can be tackled using several different approaches.

 

[MarkFL]

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.