BdMO National Secondary/Higher Secondary 2018/1

Discussion on Bangladesh Mathematical Olympiad (BdMO) National
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samiul_samin
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BdMO National Secondary/Higher Secondary 2018/1

Unread post by samiul_samin » Tue Mar 13, 2018 7:38 pm

Solve:

$x^2(2-x)^2=1+2(1-x)^2$

Where $x$ is real number.

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samiul_samin
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Re: BdMO National Higher Secondary 2018/1

Unread post by samiul_samin » Tue Mar 13, 2018 7:47 pm

Answer
1,-1,3
Solution
$x^2(2-x)^2=1+2(1-x)^2)$
$\Rightarrow 4x^2-4x^3+x^4=1+2-4x+2x^2$
$\Rightarrow x^4-4x^3+2x^2+4x-3=0$
$\Rightarrow (x-1)(x^3-3x^2-x+3)=0$
$\Rightarrow (x-1)(x-1)(x^2-2x-3)=0$
$\Rightarrow (x-1)^2 [x(x-3)+1(x-3)]=0$
$\Rightarrow (x-1)^2(x-3)(x+1)=0$
So,the answer is $\fbox {1,-1,3}$

SYED ASHFAQ TASIN
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Re: BdMO National Higher Secondary 2018/1

Unread post by SYED ASHFAQ TASIN » Wed Dec 26, 2018 9:43 pm

How do you know that x$=1$ is a solution?Trial & error?What if it had some big numbers,or fraction?
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SINAN EXPERT
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Re: BdMO National Secondary/Higher Secondary 2018/1

Unread post by SINAN EXPERT » Sat Jan 19, 2019 4:53 pm

Here is the easier way to solve the problem.
শর্তমতে,
$x^{2}\left( 2-x\right) ^{2}=1+2\left( 1-x\right) ^{2}$
বা, $x^{2}\left( x-2\right) ^{2}=1+2\left( x-1\right) ^{2}$
বা, $\left\{ x\left( x-2\right) \right\} ^{2}=1+2\left( x-1\right) ^{2}$
বা, $\left\{ (x-1+1)\left( x-1-1\right) \right\} ^{2}=1+2\left( x-1\right) ^{2}$
বা, $\left\{ \left( x-1\right) ^{2}-1^{2}\right\} ^{2}=1+2\left( x-1\right) ^{2}$
বা, $\left( x-1\right) ^{4}-2\left( x-1\right) ^{2}+1=1+2\left( x-1\right) ^{2}$
বা, $y^{2}-4y=0$ $[y=x-1$ $ধরে]$

Now you can easily solve the equation.

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samiul_samin
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Re: BdMO National Higher Secondary 2018/1

Unread post by samiul_samin » Thu Jan 31, 2019 11:15 pm

SYED ASHFAQ TASIN wrote:
Wed Dec 26, 2018 9:43 pm
How do you know that x$=1$ is a solution?Trial & error?What if it had some big numbers,or fraction?
I used trial and error to get $x=1$.

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