11(1)It's simply -1
8 Answers
622) is an RMO question. if i remember correctly, prophet sir had done this here long back..
x=r cos theta, y = r sin theta kill that one.
1708\hspace{-16}5x+\frac{5x}{x^2+y^2}=12..............................(1)\\\\ 5y-\frac{5y}{x^2+y^2}=4................................(2)\times i\\\\ $5\left(x+iy\right)+\displaystyle \frac{5}{x^2+y^2}.\left(x-iy\right)=12+4i$\\\\ $Now Let $z=x+iy$ and $\bar{z}=x-iy\;,$ So $z.\bar{z}=x^2+y^2$\\\\ $5z+\displaystyle \frac{5\bar{z}}{z.\bar{z}}=12+4i$\\\\ $5z+\frac{5}{z}=12+4i$\\\\ $5z^2-(12+4i).z+5=0$\\\\ So $z=\frac{2}{5}+i.(\frac{-1}{5})\Leftrightarrow (x,y)=\left(\frac{2}{5}\;,\frac{-1}{5}\right)$\\\\ and $z=2+i\Leftrightarrow (x,y)=(2,1)$\\\\
1708Nishant Sir I will try it using Trig. Substution.
@Shambit explanation for answer.
Thanks
11x^3+6x^2+12x+7=2.(2x+2)^(1/3)
=>(x+1)(x^2+5x+7)=2.2^(1/3)(x+1)
So, x=-1 satisfies.
This is the integer solution. There are 2 real solutions. You can check that by drawing the graph. the other one is close to -0.57.
21This RMO problem has been copied from 1996 Vietnam Olympiad problem (not exactly though). The solution posted by man111 was the desired solution. :)
3@sambit.
i think it should have been.
2.21/3.(x+1)1/3 = 24/3.(x+1)1/3
instead of 2.21/3(x+1)
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