MHB What is the value for $a+b$ in the Polynomial Challenge VI?

[anemone]

If $a,\,b$ are the two largest real roots of the polynomial $f(x)=3x^3-17x+5\sqrt{6}$, and their sum can be expressed as $\dfrac{\sqrt{m}+\sqrt{n}}{k}$ for positive integers $m,\,n,\,k$, find the value for $a+b$.

 

[anemone]

Hint:

Spoiler

Try to think of a way to get rid of the $\sqrt{6}$ so that we could make full use of the rational root theorem to this hard challenge.

 

[Euge]

If $a,\,b$ are the two largest real roots of the polynomial $f(x)=3x^3-17x+5\sqrt{6}$, and their sum can be expressed as $\dfrac{\sqrt{m}+\sqrt{n}}{k}$ for positive integers $m,\,n,\,k$, find the value for $a+b$.

Spoiler

Since

$\displaystyle f(x\sqrt{6}) = \sqrt{6}(18x^3 - 17x + 5) = \sqrt{6}(3x - 1)(6x^2 + 2x - 5)$

has roots $\frac{1}{3}$, $\frac{-1 + \sqrt{31}}{6}$, and $\frac{-1 - \sqrt{31}}{6}$, the sum of the two largest roots is

$\displaystyle \frac{1}{3} + \frac{-1 + \sqrt{31}}{6} = \frac{1 + \sqrt{31}}{6}$.

Hence

$a + b = \sqrt{6} \cdot \frac{1+\sqrt{31}}{6} = \frac{\sqrt{6} + \sqrt{186}}{6}$.

 

[Albert1]

If $a,\,b$ are the two largest real roots of the polynomial $f(x)=3x^3-17x+5\sqrt{6}$, and their sum can be expressed as $\dfrac{\sqrt{m}+\sqrt{n}}{k}$ for positive integers $m,\,n,\,k$, find the value for $a+b$.

my solution:

Spoiler

let $f(x)=(3x-p)(x-b)(x-c)=3x^3-17x+5\sqrt 6$
here $a=\dfrac {p}{3},b,c $ are three roots of $f(x)$
and $a>0,b>0,c<0$
for $a+b=\dfrac {p}{3}+b=\dfrac {\sqrt m+\sqrt n}{k}$
if $b=\dfrac {-\sqrt m+\sqrt n}{k}$ then $c=\dfrac {-\sqrt m-\sqrt n}{k}$
by comparing and observation we get $k=6$, and $ a=\dfrac {\sqrt m}{3}$
since $abc=\dfrac{\sqrt m}{3}(\dfrac {-\sqrt m+\sqrt n}{6})(\dfrac {-\sqrt m-\sqrt n}{6})$=$\dfrac{\sqrt m(n-m)}{108}=\dfrac {5\sqrt 6}{3}=\dfrac {180\sqrt 6}{108}$
$\therefore m=6,n=186$
$a+b=\dfrac{\sqrt 6+\sqrt {186}}{6}$

 

[anemone]

Well done, Euge for solving this problem in such an elegant way!

Bravo, Albert for showing us another route to tackle the problem!

Thank you both for participating.:)

 

[Albert1]

Spoiler

Since

$\displaystyle f(x\sqrt{6}) = \sqrt{6}(18x^3 - 17x + 5) = \sqrt{6}(3x - 1)(6x^2 + 2x - 5)$

has roots $\frac{1}{3}$, $\frac{-1 + \sqrt{31}}{6}$, and $\frac{-1 - \sqrt{31}}{6}$, the sum of the two largest roots is

$\displaystyle \frac{1}{3} + \frac{-1 + \sqrt{31}}{6} = \frac{1 + \sqrt{31}}{6}$.

Hence

$a + b = \frac{1}{\sqrt{6}} \cdot \frac{1+\sqrt{31}}{6} = \frac{\sqrt{6}}{6} \cdot \frac{1 + \sqrt{31}}{6} = \frac{\sqrt{6} + \sqrt{186}}{36}$.

very good soluion ,but I think ,it should be:
$a + b ={\sqrt{6}} \cdot \frac{1+\sqrt{31}}{6}= \frac{\sqrt{6} + \sqrt{186}}{6}$

 

[Euge]

very good soluion ,but I think ,it should be:
$a + b ={\sqrt{6}} \cdot \frac{1+\sqrt{31}}{6}= \frac{\sqrt{6} + \sqrt{186}}{6}$

Yes, you're right. I have made the correction. Thank you.

 

[anemone]

very good soluion ,but I think ,it should be:
$a + b ={\sqrt{6}} \cdot \frac{1+\sqrt{31}}{6}= \frac{\sqrt{6} + \sqrt{186}}{6}$

I guess I owe you a thank because as the OP to this particular challenge, I should be the one who checked for the accuracy, typos, etc from the submissions of our members...

But I didn't do that very well, because whenever I read a solution, I would focus more on seeing the tactic or skill that one used and if the poster used the "correct" method, then I don't really read for the whole post, I know this is a very bad ethic and I would change to become a better me from now on.