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BdMO National Secondary/Higher Secondary 2018/1
Posted: Tue Mar 13, 2018 7:38 pm
by samiul_samin
Solve:
$x^2(2-x)^2=1+2(1-x)^2$
Where $x$ is real number.
Re: BdMO National Higher Secondary 2018/1
Posted: Tue Mar 13, 2018 7:47 pm
by samiul_samin
Re: BdMO National Higher Secondary 2018/1
Posted: Wed Dec 26, 2018 9:43 pm
by SYED ASHFAQ TASIN
How do you know that x$=1$ is a solution?Trial & error?What if it had some big numbers,or fraction?
Re: BdMO National Secondary/Higher Secondary 2018/1
Posted: Sat Jan 19, 2019 4:53 pm
by SINAN EXPERT
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.
Re: BdMO National Higher Secondary 2018/1
Posted: Thu Jan 31, 2019 11:15 pm
by samiul_samin
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 $x=1$.