BDMO 2017 National round Secondary 3
- ahmedittihad
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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 $a+b+4c$.
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Re: BDMO 2017 National round Secondary 3
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 $x^2$.
Re: BDMO 2017 National round Secondary 3
Stuck at last step. Please help if anyone can solve
- ahmedittihad
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- Joined:Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 3
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, $a+b+4c=(n-10)+(3n+3)+4(-n)=-7$
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.
- ahmedittihad
- Posts:181
- Joined:Mon Mar 28, 2016 6:21 pm
Re: BDMO 2017 National round Secondary 3
$b=-3c+3$ So you made a mistake around the last part.
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Re: BDMO 2017 National round Secondary 3
Sorry,I was totally wrong.samiul_samin wrote: ↑Fri Feb 02, 2018 2:16 ama=-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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- Posts:1007
- Joined:Sat Dec 09, 2017 1:32 pm
Re: BDMO 2017 National round Secondary 3
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 .
$x^2+3x-1=0
\Rightarrow x^2=1-3x
\Rightarrow x^4=1-6x+9x^2
\Rightarrow x^4-9x^2+6x-1=0$
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 .