BDMO 2017 National round Secondary 3

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ahmedittihad
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BDMO 2017 National round Secondary 3

Unread post by ahmedittihad » Fri Feb 10, 2017 8:49 pm

The roots of the equation $x^2 +3x -1=0$ are also roots of the quartic equation $x^4 +ax^2 +bx +c=0$. Find the value of $a+b+4c$.
Frankly, my dear, I don't give a damn.

samiul_samin
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Re: BDMO 2017 National round Secondary 3

Unread post by samiul_samin » Fri Feb 02, 2018 2:16 am

a=-9,b=4,c=-1 .So, the desired answer is -7.We can easyliy get these values by squaring both side of the first equation when in the left side there will be only $x^2$.

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Tasnood
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Re: BDMO 2017 National round Secondary 3

Unread post by Tasnood » Mon Feb 05, 2018 10:18 pm

Stuck at last step. Please help if anyone can solve
Let $r_1$ and $r_2$ be the roots of the equation $x^2+3x-1=0$. So, we can write:
$r_1+r_2=\frac{-b}{a}=\frac{-3}{1}=-3$...1|)
$r_1r_2=\frac{c}{a}=\frac{-1}{1}=-1$...(2)

We can write the quartic equation in this way: $x^4+dx^3+ax^2+bx+c=0$ where, $d=0$
So$, d=r_1+r_2+r_3+r_4$
$\Rightarrow 0=-3+r_3+r_4$ [From (1)]
$\Rightarrow r_3+r_4=3$...(3)

So$, a=r_1r_2+r_3r_4+r_1r_3+r_1r_4+r_2r_3+r_2r_4$
$\Rightarrow a=-1-c+{r_1(r_3+r_4)+r_2(r_3+r_4)}$
$\Rightarrow a=-1-c+(r_1+r_2)(r_3+r_4)$
$\Rightarrow a=-1-c+(-3 \times 3)$[From (1) and (3)]
$\Rightarrow a=-10-c$
$\Rightarrow a+c=-10$

$c=r_1r_2r_3r_4=r_3r_4(-1)=-r_3r_4$ So, $r_3r_4=-c$

So$, b=r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4$
$\Rightarrow b=r_1r_2(r_3+r_4)+r_3r_4(r_1+r_2)$
$\Rightarrow b=(-1 \times 3){(-c) \times (-3)}$ [From (2) and (3)]
$\Rightarrow b=-3+3c$

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ahmedittihad
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Re: BDMO 2017 National round Secondary 3

Unread post by ahmedittihad » Tue Feb 06, 2018 1:36 pm

I'll just give my solution. The polynomial $x^2+3x-1$ divides $x^4+ax^2+bx+c$. So, there exists a polynomial $P(x)$ such that $x^2+3x-1 \times P(x) = x^4+ax^2+bx+c$. Obviously, $P(x)$ is a monic quadratic. Let $P(x)= x^2+mx+n$.

Then, $x^2+3x-1 \times P(x) = (x^2+3x-1) \times (x^2+mx+n) = x^4+(m+3)x^3+(n-1+3m)x^2+(3n-m)x-n $

Now, $x^4+(m+3)x^3+(n-1+3m)x^2+(3n-m)x-n = x^4+ax^2+bx+c $

Or, $m+3=0$, implies $m=-3$. So, $a=n-1-9=n-10$, $b=3n-m=3n+3$ and $c=-n$

So, $a+b+4c=(n-10)+(3n+3)+4(-n)=-7$
Frankly, my dear, I don't give a damn.

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ahmedittihad
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Re: BDMO 2017 National round Secondary 3

Unread post by ahmedittihad » Tue Feb 06, 2018 1:40 pm

Tasnood wrote:
Mon Feb 05, 2018 10:18 pm
Stuck at last step. Please help if anyone can solve
Let $r_1$ and $r_2$ be the roots of the equation $x^2+3x-1=0$. So, we can write:
$r_1+r_2=\frac{-b}{a}=\frac{-3}{1}=-3$...1|)
$r_1r_2=\frac{c}{a}=\frac{-1}{1}=-1$...(2)

We can write the quartic equation in this way: $x^4+dx^3+ax^2+bx+c=0$ where, $d=0$
So$, d=r_1+r_2+r_3+r_4$
$\Rightarrow 0=-3+r_3+r_4$ [From (1)]
$\Rightarrow r_3+r_4=3$...(3)

So$, a=r_1r_2+r_3r_4+r_1r_3+r_1r_4+r_2r_3+r_2r_4$
$\Rightarrow a=-1-c+{r_1(r_3+r_4)+r_2(r_3+r_4)}$
$\Rightarrow a=-1-c+(r_1+r_2)(r_3+r_4)$
$\Rightarrow a=-1-c+(-3 \times 3)$[From (1) and (3)]
$\Rightarrow a=-10-c$
$\Rightarrow a+c=-10$

$c=r_1r_2r_3r_4=r_3r_4(-1)=-r_3r_4$ So, $r_3r_4=-c$

So$, b=r_1r_2r_3+r_1r_2r_4+r_1r_3r_4+r_2r_3r_4$
$\Rightarrow b=r_1r_2(r_3+r_4)+r_3r_4(r_1+r_2)$
$\Rightarrow b=(-1 \times 3){(-c) \times (-3)}$ [From (2) and (3)]
$\Rightarrow b=-3+3c$
$b=-3c+3$ So you made a mistake around the last part.
Frankly, my dear, I don't give a damn.

samiul_samin
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Re: BDMO 2017 National round Secondary 3

Unread post by samiul_samin » Tue Feb 13, 2018 8:07 pm

samiul_samin wrote:
Fri Feb 02, 2018 2:16 am
a=-9,b=4,c=-1 .So, the desired answer is -7.We can easyliy get these values by squaring both side of the first equation when in the left side there will be only $x^2$.
Sorry,I was totally wrong.

samiul_samin
Posts:1007
Joined:Sat Dec 09, 2017 1:32 pm

Re: BDMO 2017 National round Secondary 3

Unread post by samiul_samin » Wed Feb 14, 2018 12:18 pm

It is given that

$x^2+3x-1=0
\Rightarrow x^2=1-3x
\Rightarrow x^4=1-6x+9x^2
\Rightarrow x^4-9x^2+6x-1=0$
By equating coefficients we can get that
$a=-9,
b=6,
c=-1$

So,$a+b+4c=-7

This solution is by my friend Afser Adil Olin.Is there any gap in this solution?Thanks .

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